Systems of Equations

A system of equations is a collection of two or more equations with a common set of variables. The solution to a system of equations is the set of values for the variables that satisfy all of the equations simultaneously. Systems of equations are widely used in various fields such as physics, engineering, economics, and many other disciplines.

There are different methods to solve systems of equations, including:

1. Substitution Method: Solving one equation for one variable and then substituting that value into the other equation.
2. Elimination Method: Adding or subtracting the equations to eliminate one of the variables.
3. Graphical Method: Plotting both equations on a graph and finding their intersection point.
4. Matrix Method (for larger systems): Using matrices and matrix operations to solve systems of equations.

---

Types of Systems of Equations

1. Consistent Systems: These systems have at least one solution.
   • Independent Systems: There is exactly one solution.
   • Dependent Systems: There are infinitely many solutions.
2. Inconsistent Systems: These systems have no solutions.

---

Solving Systems of Equations: Step-by-Step Explanation

1. Substitution Method

The substitution method involves solving one of the equations for one variable and substituting that expression into the other equation.

Steps:

1. Solve one equation for one variable.
2. Substitute that expression into the other equation.
3. Solve the new equation.
4. Substitute the solution back into one of the original equations to find the other variable.

Example:

Solve the system of equations:
$x+y=10$
$2x–y=4$

Step 1: Solve the first equation for $x$:
$x=10–y$

Step 2: Substitute this expression into the second equation:
$2(10–y)–y=4$
$20–2y–y=4$
$20–3y=4$
$-3y=4–20$
$-3y=-16$
$y=\frac{-16}{-3}$
$y=\frac{16}{3}$

Step 3: Substitute $y=\frac{16}{3}$ into $x=10–y$:
$x=10–\frac{16}{3}$
$x=\frac{30}{3}–\frac{16}{3}$
$x=\frac{14}{3}$

Thus, the solution is $x=\frac{14}{3}$ and $y=\frac{16}{3}$.

---

2. Elimination Method

The elimination method involves adding or subtracting the equations to eliminate one variable.

Steps:

1. Multiply one or both equations to make the coefficients of one variable the same.
2. Add or subtract the equations to eliminate one variable.
3. Solve for the remaining variable.
4. Substitute the solution back into one of the original equations to find the other variable.

Example:

Solve the system of equations:
$3x+2y=7$
$4x–2y=8$

Step 1: Add the two equations to eliminate $y$:
$3x+2y+4x–2y=7+8$
$7x=15$
$x=\frac{15}{7}$

Step 2: Substitute $x=\frac{15}{7}$ into the first equation:
$3\left(\frac{15}{7}\right)+2y=7$
$\frac{45}{7}+2y=7$
$2y=7–\frac{45}{7}$
$2y=\frac{49}{7}–\frac{45}{7}$
$2y=\frac{4}{7}$
$y=\frac{2}{7}$

Thus, the solution is $x=\frac{15}{7}$ and $y=\frac{2}{7}$.

---

3. Graphical Method

The graphical method involves graphing both equations and identifying the point where the lines intersect.

Example:

Solve the system of equations:
$y=2x+1$
$y=-x+4$

Step 1: Graph both equations.

Step 2: Find the point of intersection.
The point of intersection is $(1,3)$.

Thus, the solution is $x=1$ and $y=3$.

---

100 Examples of Systems of Equations

---

Examples 1 to 10

---

Example 1:
Solve the system of equations:
$x+y=6$
$2x–y=3$

Solution:
From the first equation, solve for $y$:
$y=6–x$.

Substitute into the second equation:
$2x–(6–x)=3$
$2x–6+x=3$
$3x=9$
$x=3$.

Substitute $x=3$ into $y=6–x$:
$y=6–3=3$.

Thus, $x=3$ and $y=3$.

---

Example 2:
Solve the system of equations:
$4x+3y=25$
$5x–2y=15$

Solution using elimination:
Multiply the first equation by 2 and the second equation by 3:
$2(4x+3y=25)\to 8x+6y=50$
$3(5x–2y=15)\to 15x–6y=45$.

Now add both equations:
$(8x+6y)+(15x–6y)=50+45$
$23x=95$
$x=\frac{95}{23}=5$.

Substitute $x=5$ into the first equation:
$4(5)+3y=25$
$20+3y=25$
$3y=25–20$
$3y=5$
$y=\frac{5}{3}$.

Thus, $x=5$ and $y=\frac{5}{3}$.

---

Example 3:
Solve the system of equations:
$2x–y=7$
$3x+4y=1$

Solution using substitution:
From the first equation, solve for $y$:
$y=2x–7$.

Substitute into the second equation:
$3x+4(2x–7)=1$
$3x+8x–28=1$
$11x–28=1$
$11x=1+28$
$11x=29$
$x=\frac{29}{11}$.

Substitute $x=\frac{29}{11}$ into $y=2x–7$:
$y=2\left(\frac{29}{11}\right)–7$
$y=\frac{58}{11}–7$
$y=\frac{58}{11}–\frac{77}{11}$
$y=\frac{-19}{11}$.

Thus, $x=\frac{29}{11}$ and $y=\frac{-19}{11}$.

---

Example 4:
Solve the system of equations:
$x–2y=1$
$3x+y=12$

Solution using substitution:
From the first equation, solve for $x$:
$x=2y+1$.

Substitute into the second equation:
$3(2y+1)+y=12$
$6y+3+y=12$
$7y+3=12$
$7y=12–3$
$7y=9$
$y=\frac{9}{7}$.

Substitute $y=\frac{9}{7}$ into $x=2y+1$:
$x=2\left(\frac{9}{7}\right)+1$
$x=\frac{18}{7}+1$
$x=\frac{18}{7}+\frac{7}{7}$
$x=\frac{25}{7}$.

Thus, $x=\frac{25}{7}$ and $y=\frac{9}{7}$.

---

Example 5:
Solve the system of equations:
$4x–y=10$
$2x+3y=12$

Solution using substitution:
From the first equation, solve for $y$:
$y=4x–10$.

Substitute into the second equation:
$2x+3(4x–10)=12$
$2x+12x–30=12$
$14x–30=12$
$14x=12+30$
$14x=42$
$x=\frac{42}{14}$
$x=3$.

Substitute $x=3$ into $y=4x–10$:
$y=4(3)–10$
$y=12–10$
$y=2$.

Thus, $x=3$ and $y=2$.

---

Example 6:
Solve the system of equations:
$5x+3y=29$
$2x–4y=3$

Solution using elimination:
Multiply the first equation by 4 and the second equation by 3:
$4(5x+3y=29)\to 20x+12y=116$
$3(2x–4y=3)\to 6x–12y=9$.

Add the equations:
$(20x+12y)+(6x–12y)=116+9$
$26x=125$
$x=\frac{125}{26}$.

Substitute $x=\frac{125}{26}$ into the first equation:
$5\left(\frac{125}{26}\right)+3y=29$
$\frac{625}{26}+3y=29$
$3y=29–\frac{625}{26}$
$y=\frac{754}{26}–\frac{625}{26}$
$y=\frac{129}{26}$.

Thus, $x=\frac{125}{26}$ and $y=\frac{129}{26}$.

---

Example 7:
Solve the system of equations:
$x+2y=9$
$2x–y=5$

Solution using substitution:
From the first equation, solve for $x$:
$x=9–2y$.

Substitute into the second equation:
$2(9–2y)–y=5$
$18–4y–y=5$
$18–5y=5$
$-5y=5–18$
$-5y=-13$
$y=\frac{-13}{-5}=\frac{13}{5}$.

Substitute $y=\frac{13}{5}$ into $x=9–2y$:
$x=9–2\left(\frac{13}{5}\right)$
$x=9–\frac{26}{5}$
$x=\frac{45}{5}–\frac{26}{5}$
$x=\frac{19}{5}$.

Thus, $x=\frac{19}{5}$ and $y=\frac{13}{5}$.

---

Example 8:
Solve the system of equations:
$6x–y=17$
$3x+2y=14$

Solution using substitution:
From the first equation, solve for $y$:
$y=6x–17$.

Substitute into the second equation:
$3x+2(6x–17)=14$
$3x+12x–34=14$
$15x–34=14$
$15x=14+34$
$15x=48$
$x=\frac{48}{15}$
$x=\frac{16}{5}$.

Substitute $x=\frac{16}{5}$ into $y=6x–17$:
$y=6\left(\frac{16}{5}\right)–17$
$y=\frac{96}{5}–17$
$y=\frac{96}{5}–\frac{85}{5}$
$y=\frac{11}{5}$.

Thus, $x=\frac{16}{5}$ and $y=\frac{11}{5}$.

---

Example 9:
Solve the system of equations:
$4x+5y=23$
$2x–3y=-7$

Solution using elimination:
Multiply the first equation by 3 and the second equation by 5:
$3(4x+5y=23)\to 12x+15y=69$
$5(2x–3y=-7)\to 10x–15y=-35$.

Now add both equations:
$(12x+15y)+(10x–15y)=69–35$
$22x=34$
$x=\frac{34}{22}=\frac{17}{11}$.

Substitute $x=\frac{17}{11}$ into the first equation:
$4\left(\frac{17}{11}\right)+5y=23$
$\frac{68}{11}+5y=23$
$5y=23–\frac{68}{11}$
$y=\frac{253}{11}–\frac{68}{11}$
$y=\frac{185}{11}$.

Thus, $x=\frac{17}{11}$ and $y=\frac{185}{11}$.

---

Example 10:
Solve the system of equations:
$5x+7y=41$
$2x–4y=6$

Solution using substitution:
From the first equation, solve for $x$:
$x=\frac{41–7y}{5}$.

Substitute into the second equation:
$2\left(\frac{41–7y}{5}\right)–4y=6$
$\frac{82–14y}{5}–4y=6$
Multiply by 5 to eliminate the fraction:
$82–14y–20y=30$
$82–34y=30$
$-34y=30–82$
$-34y=-52$
$y=\frac{-52}{-34}=\frac{26}{17}$.

Substitute $y=\frac{26}{17}$ into $x=\frac{41–7y}{5}$:
$x=\frac{41–7\left(\frac{26}{17}\right)}{5}$
$x=\frac{41–\frac{182}{17}}{5}$
$x=\frac{\frac{697}{17}–\frac{182}{17}}{5}$
$x=\frac{515}{17\times 5}$
$x=\frac{103}{17}$.

Thus, $x=\frac{103}{17}$ and $y=\frac{26}{17}$.

---

Examples 11 to 20

---

Example 11:
Solve the system of equations:
$2x+3y=13$
$4x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=4x–7$.

Substitute into the first equation:
$2x+3(4x–7)=13$
$2x+12x–21=13$
$14x–21=13$
$14x=13+21$
$14x=34$
$x=\frac{34}{14}=\frac{17}{7}$.

Substitute $x=\frac{17}{7}$ into $y=4x–7$:
$y=4\left(\frac{17}{7}\right)–7$
$y=\frac{68}{7}–7$
$y=\frac{68}{7}–\frac{49}{7}$
$y=\frac{19}{7}$.

Thus, $x=\frac{17}{7}$ and $y=\frac{19}{7}$.

---

Example 12:
Solve the system of equations:
$3x+y=10$
$2x–y=5$

Solution using elimination:
Add both equations to eliminate $y$:
$(3x+y)+(2x–y)=10+5$
$5x=15$
$x=\frac{15}{5}=3$.

Substitute $x=3$ into the first equation:
$3(3)+y=10$
$9+y=10$
$y=10–9$
$y=1$.

Thus, $x=3$ and $y=1$.

---

Example 13:
Solve the system of equations:
$x–4y=-7$
$5x+3y=18$

Solution using substitution:
From the first equation, solve for $x$:
$x=4y–7$.

Substitute into the second equation:
$5(4y–7)+3y=18$
$20y–35+3y=18$
$23y–35=18$
$23y=18+35$
$23y=53$
$y=\frac{53}{23}$.

Substitute $y=\frac{53}{23}$ into $x=4y–7$:
$x=4\left(\frac{53}{23}\right)–7$
$x=\frac{212}{23}–7$
$x=\frac{212}{23}–\frac{161}{23}$
$x=\frac{51}{23}$.

Thus, $x=\frac{51}{23}$ and $y=\frac{53}{23}$.

---

Example 14:
Solve the system of equations:
$7x+2y=16$
$3x–5y=-11$

Solution using elimination:
Multiply the first equation by 5 and the second equation by 2:
$5(7x+2y=16)\to 35x+10y=80$
$2(3x–5y=-11)\to 6x–10y=-22$.

Now add both equations:
$(35x+10y)+(6x–10y)=80+(-22)$
$41x=58$
$x=\frac{58}{41}$.

Substitute $x=\frac{58}{41}$ into the first equation:
$7\left(\frac{58}{41}\right)+2y=16$
$\frac{406}{41}+2y=16$
$2y=16–\frac{406}{41}$
$y=\frac{656}{41}–\frac{406}{41}$
$y=\frac{250}{41}$.

Thus, $x=\frac{58}{41}$ and $y=\frac{250}{41}$.

---

Example 15:
Solve the system of equations:
$6x+7y=22$
$x–5y=3$

Solution using substitution:
From the second equation, solve for $x$:
$x=5y+3$.

Substitute into the first equation:
$6(5y+3)+7y=22$
$30y+18+7y=22$
$37y+18=22$
$37y=22–18$
$37y=4$
$y=\frac{4}{37}$.

Substitute $y=\frac{4}{37}$ into $x=5y+3$:
$x=5\left(\frac{4}{37}\right)+3$
$x=\frac{20}{37}+3$
$x=\frac{20}{37}+\frac{111}{37}$
$x=\frac{131}{37}$.

Thus, $x=\frac{131}{37}$ and $y=\frac{4}{37}$.

---

Example 16:
Solve the system of equations:
$4x–2y=8$
$3x+y=9$

Solution using substitution:
From the second equation, solve for $y$:
$y=9–3x$.

Substitute into the first equation:
$4x–2(9–3x)=8$
$4x–18+6x=8$
$10x–18=8$
$10x=8+18$
$10x=26$
$x=\frac{26}{10}=\frac{13}{5}$.

Substitute $x=\frac{13}{5}$ into $y=9–3x$:
$y=9–3\left(\frac{13}{5}\right)$
$y=9–\frac{39}{5}$
$y=\frac{45}{5}–\frac{39}{5}$
$y=\frac{6}{5}$.

Thus, $x=\frac{13}{5}$ and $y=\frac{6}{5}$.

---

Example 17:
Solve the system of equations:
$2x+5y=18$
$4x–3y=2$

Solution using elimination:
Multiply the first equation by 3 and the second equation by 5:
$3(2x+5y=18)\to 6x+15y=54$
$5(4x–3y=2)\to 20x–15y=10$.

Now add both equations:
$(6x+15y)+(20x–15y)=54+10$
$26x=64$
$x=\frac{64}{26}=\frac{32}{13}$.

Substitute $x=\frac{32}{13}$ into the first equation:
$2\left(\frac{32}{13}\right)+5y=18$
$\frac{64}{13}+5y=18$
$5y=18–\frac{64}{13}$
$y=\frac{234}{13}–\frac{64}{13}$
$y=\frac{170}{13}$.

Thus, $x=\frac{32}{13}$ and $y=\frac{170}{13}$.

---

Example 18:
Solve the system of equations:
$5x–y=8$
$3x+4y=24$

Solution using substitution:
From the first equation, solve for $y$:
$y=5x–8$.

Substitute into the second equation:
$3x+4(5x–8)=24$
$3x+20x–32=24$
$23x–32=24$
$23x=24+32$
$23x=56$
$x=\frac{56}{23}$.

Substitute $x=\frac{56}{23}$ into $y=5x–8$:
$y=5\left(\frac{56}{23}\right)–8$
$y=\frac{280}{23}–8$
$y=\frac{280}{23}–\frac{184}{23}$
$y=\frac{96}{23}$.

Thus, $x=\frac{56}{23}$ and $y=\frac{96}{23}$.

---

Example 19:
Solve the system of equations:
$x+2y=12$
$2x–3y=-5$

Solution using substitution:
From the first equation, solve for $x$:
$x=12–2y$.

Substitute into the second equation:
$2(12–2y)–3y=-5$
$24–4y–3y=-5$
$24–7y=-5$
$-7y=-5–24$
$-7y=-29$
$y=\frac{-29}{-7}=\frac{29}{7}$.

Substitute $y=\frac{29}{7}$ into $x=12–2y$:
$x=12–2\left(\frac{29}{7}\right)$
$x=12–\frac{58}{7}$
$x=\frac{84}{7}–\frac{58}{7}$
$x=\frac{26}{7}$.

Thus, $x=\frac{26}{7}$ and $y=\frac{29}{7}$.

---

Example 20:
Solve the system of equations:
$2x+6y=16$
$3x–2y=5$

Solution using substitution:
From the first equation, solve for $x$:
$x=\frac{16–6y}{2}$
$x=8–3y$.

Substitute into the second equation:
$3(8–3y)–2y=5$
$24–9y–2y=5$
$24–11y=5$
$-11y=5–24$
$-11y=-19$
$y=\frac{-19}{-11}=\frac{19}{11}$.

Substitute $y=\frac{19}{11}$ into $x=8–3y$:
$x=8–3\left(\frac{19}{11}\right)$
$x=8–\frac{57}{11}$
$x=\frac{88}{11}–\frac{57}{11}$
$x=\frac{31}{11}$.

Thus, $x=\frac{31}{11}$ and $y=\frac{19}{11}$.

---

Examples 21 to 30

---

Example 21:
Solve the system of equations:
$4x–y=9$
$2x+3y=17$

Solution using substitution:
From the first equation, solve for $y$:
$y=4x–9$.

Substitute into the second equation:
$2x+3(4x–9)=17$
$2x+12x–27=17$
$14x–27=17$
$14x=17+27$
$14x=44$
$x=\frac{44}{14}=\frac{22}{7}$.

Substitute $x=\frac{22}{7}$ into $y=4x–9$:
$y=4\left(\frac{22}{7}\right)–9$
$y=\frac{88}{7}–9$
$y=\frac{88}{7}–\frac{63}{7}$
$y=\frac{25}{7}$.

Thus, $x=\frac{22}{7}$ and $y=\frac{25}{7}$.

---

Example 22:
Solve the system of equations:
$5x+2y=11$
$3x–y=4$

Solution using substitution:
From the second equation, solve for $y$:
$y=3x–4$.

Substitute into the first equation:
$5x+2(3x–4)=11$
$5x+6x–8=11$
$11x–8=11$
$11x=11+8$
$11x=19$
$x=\frac{19}{11}$.

Substitute $x=\frac{19}{11}$ into $y=3x–4$:
$y=3\left(\frac{19}{11}\right)–4$
$y=\frac{57}{11}–4$
$y=\frac{57}{11}–\frac{44}{11}$
$y=\frac{13}{11}$.

Thus, $x=\frac{19}{11}$ and $y=\frac{13}{11}$.

---

Example 23:
Solve the system of equations:
$6x–y=2$
$4x+3y=12$

Solution using substitution:
From the first equation, solve for $y$:
$y=6x–2$.

Substitute into the second equation:
$4x+3(6x–2)=12$
$4x+18x–6=12$
$22x–6=12$
$22x=12+6$
$22x=18$
$x=\frac{18}{22}=\frac{9}{11}$.

Substitute $x=\frac{9}{11}$ into $y=6x–2$:
$y=6\left(\frac{9}{11}\right)–2$
$y=\frac{54}{11}–2$
$y=\frac{54}{11}–\frac{22}{11}$
$y=\frac{32}{11}$.

Thus, $x=\frac{9}{11}$ and $y=\frac{32}{11}$.

---

Example 24:
Solve the system of equations:
$3x+4y=18$
$5x–2y=7$

Solution using elimination:
Multiply the first equation by 2 and the second equation by 4 to make the coefficients of $y$ the same:
$2(3x+4y=18)\to 6x+8y=36$
$5x–2y=7\to 5x–4y=14$.

Now add both equations:
$(6x+8y)+(5x–4y)=36+14$
$11x+4y=50$
$x=\frac{50}{11}$.

Substitute $x=\frac{50}{11}$ into the first equation:
$3\left(\frac{50}{11}\right)+4y=18$
$\frac{150}{11}+4y=18$
$4y=18–\frac{150}{11}$
$y=\frac{198}{11}–\frac{150}{11}$
$y=\frac{48}{11}$.

Thus, $x=\frac{50}{11}$ and $y=\frac{48}{11}$.

---

Example 25:
Solve the system of equations:
$7x–3y=8$
$2x+5y=19$

Solution using elimination:
Multiply the first equation by 5 and the second by 3:
$5(7x–3y=8)\to 35x–15y=40$
$3(2x+5y=19)\to 6x+15y=57$.

Now add both equations:
$(35x–15y)+(6x+15y)=40+57$
$41x=97$
$x=\frac{97}{41}$.

Substitute $x=\frac{97}{41}$ into the first equation:
$7\left(\frac{97}{41}\right)–3y=8$
$\frac{679}{41}–3y=8$
$3y=8–\frac{679}{41}$
$y=\frac{328}{41}–\frac{679}{41}$
$y=\frac{-351}{41}$.

Thus, $x=\frac{97}{41}$ and $y=\frac{-351}{41}$.

---

Example 26:
Solve the system of equations:
$4x–5y=1$
$3x+2y=12$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{4x–1}{5}$.

Substitute into the second equation:
$3x+2\left(\frac{4x–1}{5}\right)=12$
$3x+\frac{8x–2}{5}=12$.

Multiply by 5 to eliminate the denominator:
$5(3x+\frac{8x–2}{5})=12$
$15x+8x–2=12$
$23x–2=12$
$23x=12+2$
$23x=14$
$x=\frac{14}{23}$.

Substitute $x=\frac{14}{23}$ into $y=\frac{4x–1}{5}$:
$y=\frac{4\left(\frac{14}{23}\right)–1}{5}$
$y=\frac{\frac{56}{23}–1}{5}$
$y=\frac{56–23}{23\times 5}$
$y=\frac{33}{115}$.

Thus, $x=\frac{14}{23}$ and $y=\frac{33}{115}$.

---

Example 27:
Solve the system of equations:
$6x+7y=30$
$3x–5y=11$

Solution using elimination:
Multiply the first equation by 5 and the second equation by 7:
$5(6x+7y=30)\to 30x+35y=150$
$7(3x–5y=11)\to 21x–35y=77$.

Now add both equations:
$(30x+35y)+(21x–35y)=150+77$
$51x=227$
$x=\frac{227}{51}$.

Substitute $x=\frac{227}{51}$ into the first equation:
$6\left(\frac{227}{51}\right)+7y=30$
$\frac{1362}{51}+7y=30$
$7y=30–\frac{1362}{51}$
$y=\frac{1530}{51}–\frac{1362}{51}$
$y=\frac{168}{51}$.

Thus, $x=\frac{227}{51}$ and $y=\frac{168}{51}$.

---

Example 28:
Solve the system of equations:
$4x–6y=8$
$5x+2y=13$

Solution using substitution:
From the first equation, solve for $x$:
$x=\frac{6y+8}{4}$.

Substitute into the second equation:
$5\left(\frac{6y+8}{4}\right)+2y=13$
$\frac{5(6y+8)}{4}+2y=13$
$\frac{30y+40}{4}+2y=13$.

Multiply by 4 to eliminate the denominator:
$5(6y+8)+8y=52$
$30y+40+8y=52$
$38y+40=52$
$38y=52–40$
$38y=12$
$y=\frac{12}{38}=\frac{6}{19}$.

Substitute $y=\frac{6}{19}$ into $x=\frac{6y+8}{4}$:
$x=\frac{6\left(\frac{6}{19}\right)+8}{4}$
$x=\frac{\frac{36}{19}+8}{4}$
$x=\frac{36+152}{19\times 4}$
$x=\frac{188}{76}$
$x=\frac{47}{19}$.

Thus, $x=\frac{47}{19}$ and $y=\frac{6}{19}$.

---

Example 29:
Solve the system of equations:
$7x+2y=9$
$3x–4y=5$

Solution using elimination:
Multiply the first equation by 4 and the second equation by 2:
$4(7x+2y=9)\to 28x+8y=36$
$2(3x–4y=5)\to 6x–8y=10$.

Now add both equations:
$(28x+8y)+(6x–8y)=36+10$
$34x=46$
$x=\frac{46}{34}=\frac{23}{17}$.

Substitute $x=\frac{23}{17}$ into the first equation:
$7\left(\frac{23}{17}\right)+2y=9$
$\frac{161}{17}+2y=9$
$2y=9–\frac{161}{17}$
$y=\frac{153}{17}–\frac{161}{17}$
$y=\frac{-8}{17}$.

Thus, $x=\frac{23}{17}$ and $y=\frac{-8}{17}$.

---

Example 30:
Solve the system of equations:
$8x–3y=6$
$4x+y=10$

Solution using substitution:
From the second equation, solve for $y$:
$y=10–4x$.

Substitute into the first equation:
$8x–3(10–4x)=6$
$8x–30+12x=6$
$20x–30=6$
$20x=6+30$
$20x=36$
$x=\frac{36}{20}=\frac{9}{5}$.

Substitute $x=\frac{9}{5}$ into $y=10–4x$:
$y=10–4\left(\frac{9}{5}\right)$
$y=10–\frac{36}{5}$
$y=\frac{50}{5}–\frac{36}{5}$
$y=\frac{14}{5}$.

Thus, $x=\frac{9}{5}$ and $y=\frac{14}{5}$.

---

Examples 31 to 40

---

Example 31:
Solve the system of equations:
$3x+2y=12$
$5x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=5x–7$.

Substitute into the first equation:
$3x+2(5x–7)=12$
$3x+10x–14=12$
$13x–14=12$
$13x=12+14$
$13x=26$
$x=\frac{26}{13}=2$.

Substitute $x=2$ into $y=5x–7$:
$y=5(2)–7$
$y=10–7=3$.

Thus, $x=2$ and $y=3$.

---

Example 32:
Solve the system of equations:
$4x–y=10$
$3x+2y=14$

Solution using substitution:
From the first equation, solve for $y$:
$y=4x–10$.

Substitute into the second equation:
$3x+2(4x–10)=14$
$3x+8x–20=14$
$11x–20=14$
$11x=14+20$
$11x=34$
$x=\frac{34}{11}$.

Substitute $x=\frac{34}{11}$ into $y=4x–10$:
$y=4\left(\frac{34}{11}\right)–10$
$y=\frac{136}{11}–10$
$y=\frac{136}{11}–\frac{110}{11}$
$y=\frac{26}{11}$.

Thus, $x=\frac{34}{11}$ and $y=\frac{26}{11}$.

---

Example 33:
Solve the system of equations:
$6x+y=13$
$5x–2y=4$

Solution using substitution:
From the first equation, solve for $y$:
$y=13–6x$.

Substitute into the second equation:
$5x–2(13–6x)=4$
$5x–26+12x=4$
$17x–26=4$
$17x=4+26$
$17x=30$
$x=\frac{30}{17}$.

Substitute $x=\frac{30}{17}$ into $y=13–6x$:
$y=13–6\left(\frac{30}{17}\right)$
$y=13–\frac{180}{17}$
$y=\frac{221}{17}–\frac{180}{17}$
$y=\frac{41}{17}$.

Thus, $x=\frac{30}{17}$ and $y=\frac{41}{17}$.

---

Example 34:
Solve the system of equations:
$4x+5y=31$
$6x–y=19$

Solution using substitution:
From the second equation, solve for $y$:
$y=6x–19$.

Substitute into the first equation:
$4x+5(6x–19)=31$
$4x+30x–95=31$
$34x–95=31$
$34x=31+95$
$34x=126$
$x=\frac{126}{34}=\frac{63}{17}$.

Substitute $x=\frac{63}{17}$ into $y=6x–19$:
$y=6\left(\frac{63}{17}\right)–19$
$y=\frac{378}{17}–19$
$y=\frac{378}{17}–\frac{323}{17}$
$y=\frac{55}{17}$.

Thus, $x=\frac{63}{17}$ and $y=\frac{55}{17}$.

---

Example 35:
Solve the system of equations:
$7x–4y=2$
$5x+6y=15$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{7x–2}{4}$.

Substitute into the second equation:
$5x+6\left(\frac{7x–2}{4}\right)=15$
$5x+\frac{42x–12}{4}=15$.

Multiply by 4 to eliminate the denominator:
$4(5x+\frac{42x–12}{4})=4(15)$
$20x+42x–12=60$
$62x–12=60$
$62x=60+12$
$62x=72$
$x=\frac{72}{62}=\frac{36}{31}$.

Substitute $x=\frac{36}{31}$ into $y=\frac{7x–2}{4}$:
$y=\frac{7\left(\frac{36}{31}\right)–2}{4}$
$y=\frac{\frac{252}{31}–2}{4}$
$y=\frac{252–62}{31\times 4}$
$y=\frac{190}{124}$
$y=\frac{95}{62}$.

Thus, $x=\frac{36}{31}$ and $y=\frac{95}{62}$.

---

Example 36:
Solve the system of equations:
$3x+7y=22$
$4x–3y=6$

Solution using elimination:
Multiply the first equation by 3 and the second by 7 to eliminate $y$:
$3(3x+7y=22)\to 9x+21y=66$
$7(4x–3y=6)\to 28x–21y=42$.

Now add both equations:
$(9x+21y)+(28x–21y)=66+42$
$37x=108$
$x=\frac{108}{37}$.

Substitute $x=\frac{108}{37}$ into the first equation:
$3\left(\frac{108}{37}\right)+7y=22$
$\frac{324}{37}+7y=22$
$7y=22–\frac{324}{37}$
$y=\frac{814}{37}–\frac{324}{37}$
$y=\frac{490}{37}$.

Thus, $x=\frac{108}{37}$ and $y=\frac{490}{37}$.

---

Example 37:
Solve the system of equations:
$2x+4y=10$
$3x–2y=7$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{10–2x}{4}$.

Substitute into the second equation:
$3x–2\left(\frac{10–2x}{4}\right)=7$
$3x–\frac{20–4x}{4}=7$.

Multiply by 4 to eliminate the denominator:
$4(3x–\frac{20–4x}{4})=4(7)$
$12x–20+4x=28$
$16x–20=28$
$16x=28+20$
$16x=48$
$x=\frac{48}{16}=3$.

Substitute $x=3$ into $y=\frac{10–2x}{4}$:
$y=\frac{10–2(3)}{4}$
$y=\frac{10–6}{4}$
$y=\frac{4}{4}=1$.

Thus, $x=3$ and $y=1$.

---

Example 38:
Solve the system of equations:
$5x–4y=14$
$3x+2y=11$

Solution using elimination:
Multiply the first equation by 2 and the second by 4:
$2(5x–4y=14)\to 10x–8y=28$
$4(3x+2y=11)\to 12x+8y=44$.

Now add both equations:
$(10x–8y)+(12x+8y)=28+44$
$22x=72$
$x=\frac{72}{22}=\frac{36}{11}$.

Substitute $x=\frac{36}{11}$ into the first equation:
$5\left(\frac{36}{11}\right)–4y=14$
$\frac{180}{11}–4y=14$
$4y=14–\frac{180}{11}$
$y=\frac{154}{11}–\frac{180}{11}$
$y=\frac{-26}{11}$.

Thus, $x=\frac{36}{11}$ and $y=\frac{-26}{11}$.

---

Example 39:
Solve the system of equations:
$7x+3y=15$
$2x–5y=8$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{15–7x}{3}$.

Substitute into the second equation:
$2x–5\left(\frac{15–7x}{3}\right)=8$
$2x–\frac{75–35x}{3}=8$.

Multiply by 3 to eliminate the denominator:
$3(2x–\frac{75–35x}{3})=3(8)$
$6x–75+35x=24$
$41x–75=24$
$41x=24+75$
$41x=99$
$x=\frac{99}{41}$.

Substitute $x=\frac{99}{41}$ into $y=\frac{15–7x}{3}$:
$y=\frac{15–7\left(\frac{99}{41}\right)}{3}$
$y=\frac{15–\frac{693}{41}}{3}$
$y=\frac{615–693}{123}$
$y=\frac{-78}{123}$
$y=\frac{-26}{41}$.

Thus, $x=\frac{99}{41}$ and $y=\frac{-26}{41}$.

---

Example 40:
Solve the system of equations:
$3x–4y=11$
$5x+y=9$

Solution using substitution:
From the second equation, solve for $y$:
$y=9–5x$.

Substitute into the first equation:
$3x–4(9–5x)=11$
$3x–36+20x=11$
$23x–36=11$
$23x=11+36$
$23x=47$
$x=\frac{47}{23}$.

Substitute $x=\frac{47}{23}$ into $y=9–5x$:
$y=9–5\left(\frac{47}{23}\right)$
$y=9–\frac{235}{23}$
$y=\frac{207}{23}–\frac{235}{23}$
$y=\frac{-28}{23}$.

Thus, $x=\frac{47}{23}$ and $y=\frac{-28}{23}$.

---

Examples 41 to 50

---

Example 41:
Solve the system of equations:
$4x–5y=3$
$6x+y=9$

Solution using substitution:
From the second equation, solve for $y$:
$y=9–6x$.

Substitute into the first equation:
$4x–5(9–6x)=3$
$4x–45+30x=3$
$34x–45=3$
$34x=3+45$
$34x=48$
$x=\frac{48}{34}=\frac{24}{17}$.

Substitute $x=\frac{24}{17}$ into $y=9–6x$:
$y=9–6\left(\frac{24}{17}\right)$
$y=9–\frac{144}{17}$
$y=\frac{153}{17}–\frac{144}{17}$
$y=\frac{9}{17}$.

Thus, $x=\frac{24}{17}$ and $y=\frac{9}{17}$.

---

Example 42:
Solve the system of equations:
$3x+4y=16$
$7x–2y=10$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{16–3x}{4}$.

Substitute into the second equation:
$7x–2\left(\frac{16–3x}{4}\right)=10$
$7x–\frac{32–6x}{4}=10$.

Multiply by 4 to eliminate the denominator:
$4(7x–\frac{32–6x}{4})=4(10)$
$28x–32+6x=40$
$34x–32=40$
$34x=40+32$
$34x=72$
$x=\frac{72}{34}=\frac{36}{17}$.

Substitute $x=\frac{36}{17}$ into $y=\frac{16–3x}{4}$:
$y=\frac{16–3\left(\frac{36}{17}\right)}{4}$
$y=\frac{16–\frac{108}{17}}{4}$
$y=\frac{272–108}{68}$
$y=\frac{164}{68}=\frac{82}{34}$
$y=\frac{41}{17}$.

Thus, $x=\frac{36}{17}$ and $y=\frac{41}{17}$.

---

Example 43:
Solve the system of equations:
$5x+y=7$
$3x–4y=8$

Solution using substitution:
From the first equation, solve for $y$:
$y=7–5x$.

Substitute into the second equation:
$3x–4(7–5x)=8$
$3x–28+20x=8$
$23x–28=8$
$23x=8+28$
$23x=36$
$x=\frac{36}{23}$.

Substitute $x=\frac{36}{23}$ into $y=7–5x$:
$y=7–5\left(\frac{36}{23}\right)$
$y=7–\frac{180}{23}$
$y=\frac{161}{23}–\frac{180}{23}$
$y=\frac{-19}{23}$.

Thus, $x=\frac{36}{23}$ and $y=\frac{-19}{23}$.

---

Example 44:
Solve the system of equations:
$6x+7y=20$
$8x–5y=12$

Solution using elimination:
Multiply the first equation by 5 and the second by 7:
$5(6x+7y=20)\to 30x+35y=100$
$7(8x–5y=12)\to 56x–35y=84$.

Now add both equations:
$(30x+35y)+(56x–35y)=100+84$
$86x=184$
$x=\frac{184}{86}=\frac{92}{43}$.

Substitute $x=\frac{92}{43}$ into the first equation:
$6\left(\frac{92}{43}\right)+7y=20$
$\frac{552}{43}+7y=20$
$7y=20–\frac{552}{43}$
$y=\frac{860}{43}–\frac{552}{43}$
$y=\frac{308}{43}$.

Thus, $x=\frac{92}{43}$ and $y=\frac{308}{43}$.

---

Example 45:
Solve the system of equations:
$4x–y=11$
$2x+3y=12$

Solution using substitution:
From the first equation, solve for $y$:
$y=4x–11$.

Substitute into the second equation:
$2x+3(4x–11)=12$
$2x+12x–33=12$
$14x–33=12$
$14x=12+33$
$14x=45$
$x=\frac{45}{14}$.

Substitute $x=\frac{45}{14}$ into $y=4x–11$:
$y=4\left(\frac{45}{14}\right)–11$
$y=\frac{180}{14}–11$
$y=\frac{180}{14}–\frac{154}{14}$
$y=\frac{26}{14}=\frac{13}{7}$.

Thus, $x=\frac{45}{14}$ and $y=\frac{13}{7}$.

---

Example 46:
Solve the system of equations:
$5x–3y=9$
$3x+2y=7$

Solution using elimination:
Multiply the first equation by 2 and the second equation by 3:
$2(5x–3y=9)\to 10x–6y=18$
$3(3x+2y=7)\to 9x+6y=21$.

Now add both equations:
$(10x–6y)+(9x+6y)=18+21$
$19x=39$
$x=\frac{39}{19}$.

Substitute $x=\frac{39}{19}$ into the first equation:
$5\left(\frac{39}{19}\right)–3y=9$
$\frac{195}{19}–3y=9$
$3y=9–\frac{195}{19}$
$y=\frac{171}{19}–\frac{195}{19}$
$y=\frac{-24}{19}$.

Thus, $x=\frac{39}{19}$ and $y=\frac{-24}{19}$.

---

Example 47:
Solve the system of equations:
$6x+y=13$
$4x–2y=6$

Solution using substitution:
From the first equation, solve for $y$:
$y=13–6x$.

Substitute into the second equation:
$4x–2(13–6x)=6$
$4x–26+12x=6$
$16x–26=6$
$16x=6+26$
$16x=32$
$x=\frac{32}{16}=2$.

Substitute $x=2$ into $y=13–6x$:
$y=13–6(2)$
$y=13–12$
$y=1$.

Thus, $x=2$ and $y=1$.

---

Example 48:
Solve the system of equations:
$7x–5y=6$
$3x+4y=12$

Solution using elimination:
Multiply the first equation by 4 and the second equation by 5:
$4(7x–5y=6)\to 28x–20y=24$
$5(3x+4y=12)\to 15x+20y=60$.

Now add both equations:
$(28x–20y)+(15x+20y)=24+60$
$43x=84$
$x=\frac{84}{43}$.

Substitute $x=\frac{84}{43}$ into the first equation:
$7\left(\frac{84}{43}\right)–5y=6$
$\frac{588}{43}–5y=6$
$5y=6–\frac{588}{43}$
$y=\frac{258}{43}–\frac{588}{43}$
$y=\frac{-330}{43}$.

Thus, $x=\frac{84}{43}$ and $y=\frac{-330}{43}$.

---

Example 49:
Solve the system of equations:
$2x+3y=10$
$5x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=5x–7$.

Substitute into the first equation:
$2x+3(5x–7)=10$
$2x+15x–21=10$
$17x–21=10$
$17x=10+21$
$17x=31$
$x=\frac{31}{17}$.

Substitute $x=\frac{31}{17}$ into $y=5x–7$:
$y=5\left(\frac{31}{17}\right)–7$
$y=\frac{155}{17}–7$
$y=\frac{155}{17}–\frac{119}{17}$
$y=\frac{36}{17}$.

Thus, $x=\frac{31}{17}$ and $y=\frac{36}{17}$.

---

Example 50:
Solve the system of equations:
$4x–2y=8$
$3x+5y=14$

Solution using elimination:
Multiply the first equation by 5 and the second equation by 2:
$5(4x–2y=8)\to 20x–10y=40$
$2(3x+5y=14)\to 6x+10y=28$.

Now add both equations:
$(20x–10y)+(6x+10y)=40+28$
$26x=68$
$x=\frac{68}{26}=\frac{34}{13}$.

Substitute $x=\frac{34}{13}$ into the first equation:
$4\left(\frac{34}{13}\right)–2y=8$
$\frac{136}{13}–2y=8$
$2y=8–\frac{136}{13}$
$y=\frac{104}{13}–\frac{136}{13}$
$y=\frac{-32}{13}$.

Thus, $x=\frac{34}{13}$ and $y=\frac{-32}{13}$.

---

Examples 51 to 60

---

Example 51:
Solve the system of equations:
$5x–y=8$
$4x+2y=12$

Solution using substitution:
From the first equation, solve for $y$:
$y=5x–8$.

Substitute into the second equation:
$4x+2(5x–8)=12$
$4x+10x–16=12$
$14x–16=12$
$14x=12+16$
$14x=28$
$x=\frac{28}{14}=2$.

Substitute $x=2$ into $y=5x–8$:
$y=5(2)–8$
$y=10–8=2$.

Thus, $x=2$ and $y=2$.

---

Example 52:
Solve the system of equations:
$3x+y=7$
$2x–y=1$

Solution using elimination:
Add the two equations directly:
$(3x+y)+(2x–y)=7+1$
$5x=8$
$x=\frac{8}{5}$.

Substitute $x=\frac{8}{5}$ into the first equation:
$3\left(\frac{8}{5}\right)+y=7$
$\frac{24}{5}+y=7$
$y=7–\frac{24}{5}$
$y=\frac{35}{5}–\frac{24}{5}$
$y=\frac{11}{5}$.

Thus, $x=\frac{8}{5}$ and $y=\frac{11}{5}$.

---

Example 53:
Solve the system of equations:
$6x–y=14$
$7x+2y=25$

Solution using substitution:
From the first equation, solve for $y$:
$y=6x–14$.

Substitute into the second equation:
$7x+2(6x–14)=25$
$7x+12x–28=25$
$19x–28=25$
$19x=25+28$
$19x=53$
$x=\frac{53}{19}$.

Substitute $x=\frac{53}{19}$ into $y=6x–14$:
$y=6\left(\frac{53}{19}\right)–14$
$y=\frac{318}{19}–14$
$y=\frac{318}{19}–\frac{266}{19}$
$y=\frac{52}{19}$.

Thus, $x=\frac{53}{19}$ and $y=\frac{52}{19}$.

---

Example 54:
Solve the system of equations:
$5x+y=13$
$4x–3y=6$

Solution using substitution:
From the first equation, solve for $y$:
$y=13–5x$.

Substitute into the second equation:
$4x–3(13–5x)=6$
$4x–39+15x=6$
$19x–39=6$
$19x=6+39$
$19x=45$
$x=\frac{45}{19}$.

Substitute $x=\frac{45}{19}$ into $y=13–5x$:
$y=13–5\left(\frac{45}{19}\right)$
$y=13–\frac{225}{19}$
$y=\frac{247}{19}–\frac{225}{19}$
$y=\frac{22}{19}$.

Thus, $x=\frac{45}{19}$ and $y=\frac{22}{19}$.

---

Example 55:
Solve the system of equations:
$6x–3y=9$
$2x+y=4$

Solution using substitution:
From the second equation, solve for $y$:
$y=4–2x$.

Substitute into the first equation:
$6x–3(4–2x)=9$
$6x–12+6x=9$
$12x–12=9$
$12x=9+12$
$12x=21$
$x=\frac{21}{12}=\frac{7}{4}$.

Substitute $x=\frac{7}{4}$ into $y=4–2x$:
$y=4–2\left(\frac{7}{4}\right)$
$y=4–\frac{14}{4}$
$y=\frac{16}{4}–\frac{14}{4}$
$y=\frac{2}{4}=\frac{1}{2}$.

Thus, $x=\frac{7}{4}$ and $y=\frac{1}{2}$.

---

Example 56:
Solve the system of equations:
$4x+7y=24$
$5x–3y=9$

Solution using elimination:
Multiply the first equation by 3 and the second by 7:
$3(4x+7y=24)\to 12x+21y=72$
$7(5x–3y=9)\to 35x–21y=63$.

Now add both equations:
$(12x+21y)+(35x–21y)=72+63$
$47x=135$
$x=\frac{135}{47}$.

Substitute $x=\frac{135}{47}$ into the first equation:
$4\left(\frac{135}{47}\right)+7y=24$
$\frac{540}{47}+7y=24$
$7y=24–\frac{540}{47}$
$y=\frac{1128}{47}–\frac{540}{47}$
$y=\frac{588}{47}$.

Thus, $x=\frac{135}{47}$ and $y=\frac{588}{47}$.

---

Example 57:
Solve the system of equations:
$2x+5y=11$
$4x–3y=9$

Solution using elimination:
Multiply the first equation by 3 and the second equation by 5:
$3(2x+5y=11)\to 6x+15y=33$
$5(4x–3y=9)\to 20x–15y=45$.

Now add both equations:
$(6x+15y)+(20x–15y)=33+45$
$26x=78$
$x=\frac{78}{26}=3$.

Substitute $x=3$ into the first equation:
$2(3)+5y=11$
$6+5y=11$
$5y=11–6$
$5y=5$
$y=\frac{5}{5}=1$.

Thus, $x=3$ and $y=1$.

---

Example 58:
Solve the system of equations:
$3x–y=5$
$7x+2y=18$

Solution using substitution:
From the first equation, solve for $y$:
$y=3x–5$.

Substitute into the second equation:
$7x+2(3x–5)=18$
$7x+6x–10=18$
$13x–10=18$
$13x=18+10$
$13x=28$
$x=\frac{28}{13}$.

Substitute $x=\frac{28}{13}$ into $y=3x–5$:
$y=3\left(\frac{28}{13}\right)–5$
$y=\frac{84}{13}–5$
$y=\frac{84}{13}–\frac{65}{13}$
$y=\frac{19}{13}$.

Thus, $x=\frac{28}{13}$ and $y=\frac{19}{13}$.

---

Example 59:
Solve the system of equations:
$8x+y=17$
$2x–3y=9$

Solution using substitution:
From the first equation, solve for $y$:
$y=17–8x$.

Substitute into the second equation:
$2x–3(17–8x)=9$
$2x–51+24x=9$
$26x–51=9$
$26x=9+51$
$26x=60$
$x=\frac{60}{26}=\frac{30}{13}$.

Substitute $x=\frac{30}{13}$ into $y=17–8x$:
$y=17–8\left(\frac{30}{13}\right)$
$y=17–\frac{240}{13}$
$y=\frac{221}{13}–\frac{240}{13}$
$y=\frac{-19}{13}$.

Thus, $x=\frac{30}{13}$ and $y=\frac{-19}{13}$.

---

Example 60:
Solve the system of equations:
$5x+4y=23$
$7x–3y=13$

Solution using elimination:
Multiply the first equation by 3 and the second equation by 4:
$3(5x+4y=23)\to 15x+12y=69$
$4(7x–3y=13)\to 28x–12y=52$.

Now add both equations:
$(15x+12y)+(28x–12y)=69+52$
$43x=121$
$x=\frac{121}{43}$.

Substitute $x=\frac{121}{43}$ into the first equation:
$5\left(\frac{121}{43}\right)+4y=23$
$\frac{605}{43}+4y=23$
$4y=23–\frac{605}{43}$
$y=\frac{989}{43}–\frac{605}{43}$
$y=\frac{384}{43}$.

Thus, $x=\frac{121}{43}$ and $y=\frac{384}{43}$.

---

Examples 61 to 70

---

Example 61:
Solve the system of equations:
$3x+2y=11$
$4x–3y=13$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{11–3x}{2}$.

Substitute into the second equation:
$4x–3\left(\frac{11–3x}{2}\right)=13$
$4x–\frac{3(11–3x)}{2}=13$
$4x–\frac{33–9x}{2}=13$.

Multiply by 2 to eliminate the denominator:
$2(4x–\frac{33–9x}{2})=2(13)$
$8x–33+9x=26$
$17x–33=26$
$17x=26+33$
$17x=59$
$x=\frac{59}{17}$.

Substitute $x=\frac{59}{17}$ into $y=\frac{11–3x}{2}$:
$y=\frac{11–3\left(\frac{59}{17}\right)}{2}$
$y=\frac{11–\frac{177}{17}}{2}$
$y=\frac{187–177}{34}$
$y=\frac{10}{34}=\frac{5}{17}$.

Thus, $x=\frac{59}{17}$ and $y=\frac{5}{17}$.

---

Example 62:
Solve the system of equations:
$5x–4y=6$
$7x+3y=19$

Solution using elimination:
Multiply the first equation by 3 and the second equation by 4:
$3(5x–4y=6)\to 15x–12y=18$
$4(7x+3y=19)\to 28x+12y=76$.

Now add both equations:
$(15x–12y)+(28x+12y)=18+76$
$43x=94$
$x=\frac{94}{43}$.

Substitute $x=\frac{94}{43}$ into the first equation:
$5\left(\frac{94}{43}\right)–4y=6$
$\frac{470}{43}–4y=6$
$4y=6–\frac{470}{43}$
$y=\frac{258}{43}–\frac{470}{43}$
$y=\frac{-212}{43}$.

Thus, $x=\frac{94}{43}$ and $y=\frac{-212}{43}$.

---

Example 63:
Solve the system of equations:
$6x+y=17$
$3x–2y=7$

Solution using substitution:
From the first equation, solve for $y$:
$y=17–6x$.

Substitute into the second equation:
$3x–2(17–6x)=7$
$3x–34+12x=7$
$15x–34=7$
$15x=7+34$
$15x=41$
$x=\frac{41}{15}$.

Substitute $x=\frac{41}{15}$ into $y=17–6x$:
$y=17–6\left(\frac{41}{15}\right)$
$y=17–\frac{246}{15}$
$y=\frac{255}{15}–\frac{246}{15}$
$y=\frac{9}{15}=\frac{3}{5}$.

Thus, $x=\frac{41}{15}$ and $y=\frac{3}{5}$.

---

Example 64:
Solve the system of equations:
$4x–3y=5$
$7x+5y=19$

Solution using elimination:
Multiply the first equation by 5 and the second by 3:
$5(4x–3y=5)\to 20x–15y=25$
$3(7x+5y=19)\to 21x+15y=57$.

Now add both equations:
$(20x–15y)+(21x+15y)=25+57$
$41x=82$
$x=\frac{82}{41}=2$.

Substitute $x=2$ into the first equation:
$4(2)–3y=5$
$8–3y=5$
$-3y=5–8$
$y=\frac{-3}{-3}=1$.

Thus, $x=2$ and $y=1$.

---

Example 65:
Solve the system of equations:
$2x+5y=12$
$7x–y=8$

Solution using substitution:
From the second equation, solve for $y$:
$y=7x–8$.

Substitute into the first equation:
$2x+5(7x–8)=12$
$2x+35x–40=12$
$37x–40=12$
$37x=12+40$
$37x=52$
$x=\frac{52}{37}$.

Substitute $x=\frac{52}{37}$ into $y=7x–8$:
$y=7\left(\frac{52}{37}\right)–8$
$y=\frac{364}{37}–8$
$y=\frac{364}{37}–\frac{296}{37}$
$y=\frac{68}{37}$.

Thus, $x=\frac{52}{37}$ and $y=\frac{68}{37}$.

---

Example 66:
Solve the system of equations:
$3x–4y=16$
$5x+y=9$

Solution using substitution:
From the second equation, solve for $y$:
$y=9–5x$.

Substitute into the first equation:
$3x–4(9–5x)=16$
$3x–36+20x=16$
$23x–36=16$
$23x=16+36$
$23x=52$
$x=\frac{52}{23}$.

Substitute $x=\frac{52}{23}$ into $y=9–5x$:
$y=9–5\left(\frac{52}{23}\right)$
$y=9–\frac{260}{23}$
$y=\frac{207}{23}–\frac{260}{23}$
$y=\frac{-53}{23}$.

Thus, $x=\frac{52}{23}$ and $y=\frac{-53}{23}$.

---

Example 67:
Solve the system of equations:
$6x+y=20$
$8x–3y=14$

Solution using substitution:
From the first equation, solve for $y$:
$y=20–6x$.

Substitute into the second equation:
$8x–3(20–6x)=14$
$8x–60+18x=14$
$26x–60=14$
$26x=14+60$
$26x=74$
$x=\frac{74}{26}=\frac{37}{13}$.

Substitute $x=\frac{37}{13}$ into $y=20–6x$:
$y=20–6\left(\frac{37}{13}\right)$
$y=20–\frac{222}{13}$
$y=\frac{260}{13}–\frac{222}{13}$
$y=\frac{38}{13}$.

Thus, $x=\frac{37}{13}$ and $y=\frac{38}{13}$.

---

Example 68:
Solve the system of equations:
$2x+y=5$
$4x–3y=7$

Solution using substitution:
From the first equation, solve for $y$:
$y=5–2x$.

Substitute into the second equation:
$4x–3(5–2x)=7$
$4x–15+6x=7$
$10x–15=7$
$10x=7+15$
$10x=22$
$x=\frac{22}{10}=\frac{11}{5}$.

Substitute $x=\frac{11}{5}$ into $y=5–2x$:
$y=5–2\left(\frac{11}{5}\right)$
$y=5–\frac{22}{5}$
$y=\frac{25}{5}–\frac{22}{5}$
$y=\frac{3}{5}$.

Thus, $x=\frac{11}{5}$ and $y=\frac{3}{5}$.

---

Example 69:
Solve the system of equations:
$5x+2y=11$
$3x–y=4$

Solution using substitution:
From the second equation, solve for $y$:
$y=3x–4$.

Substitute into the first equation:
$5x+2(3x–4)=11$
$5x+6x–8=11$
$11x–8=11$
$11x=11+8$
$11x=19$
$x=\frac{19}{11}$.

Substitute $x=\frac{19}{11}$ into $y=3x–4$:
$y=3\left(\frac{19}{11}\right)–4$
$y=\frac{57}{11}–4$
$y=\frac{57}{11}–\frac{44}{11}$
$y=\frac{13}{11}$.

Thus, $x=\frac{19}{11}$ and $y=\frac{13}{11}$.

---

Example 70:
Solve the system of equations:
$4x–y=7$
$6x+3y=2$

Solution using substitution:
From the first equation, solve for $y$:
$y=4x–7$.

Substitute into the second equation:
$6x+3(4x–7)=2$
$6x+12x–21=2$
$18x–21=2$
$18x=2+21$
$18x=23$
$x=\frac{23}{18}$.

Substitute $x=\frac{23}{18}$ into $y=4x–7$:
$y=4\left(\frac{23}{18}\right)–7$
$y=\frac{92}{18}–7$
$y=\frac{92}{18}–\frac{126}{18}$
$y=\frac{-34}{18}=\frac{-17}{9}$.

Thus, $x=\frac{23}{18}$ and $y=\frac{-17}{9}$.

---

Examples 71 to 80

---

Example 71:
Solve the system of equations:
$2x+3y=8$
$4x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=4x–7$.

Substitute into the first equation:
$2x+3(4x–7)=8$
$2x+12x–21=8$
$14x–21=8$
$14x=8+21$
$14x=29$
$x=\frac{29}{14}$.

Substitute $x=\frac{29}{14}$ into $y=4x–7$:
$y=4\left(\frac{29}{14}\right)–7$
$y=\frac{116}{14}–7$
$y=\frac{116}{14}–\frac{98}{14}$
$y=\frac{18}{14}=\frac{9}{7}$.

Thus, $x=\frac{29}{14}$ and $y=\frac{9}{7}$.

---

Example 72:
Solve the system of equations:
$5x–y=10$
$3x+4y=20$

Solution using substitution:
From the first equation, solve for $y$:
$y=5x–10$.

Substitute into the second equation:
$3x+4(5x–10)=20$
$3x+20x–40=20$
$23x–40=20$
$23x=20+40$
$23x=60$
$x=\frac{60}{23}$.

Substitute $x=\frac{60}{23}$ into $y=5x–10$:
$y=5\left(\frac{60}{23}\right)–10$
$y=\frac{300}{23}–10$
$y=\frac{300}{23}–\frac{230}{23}$
$y=\frac{70}{23}$.

Thus, $x=\frac{60}{23}$ and $y=\frac{70}{23}$.

---

Example 73:
Solve the system of equations:
$6x+y=14$
$5x–2y=13$

Solution using substitution:
From the first equation, solve for $y$:
$y=14–6x$.

Substitute into the second equation:
$5x–2(14–6x)=13$
$5x–28+12x=13$
$17x–28=13$
$17x=13+28$
$17x=41$
$x=\frac{41}{17}$.

Substitute $x=\frac{41}{17}$ into $y=14–6x$:
$y=14–6\left(\frac{41}{17}\right)$
$y=14–\frac{246}{17}$
$y=\frac{238}{17}–\frac{246}{17}$
$y=\frac{-8}{17}$.

Thus, $x=\frac{41}{17}$ and $y=\frac{-8}{17}$.

---

Example 74:
Solve the system of equations:
$2x+4y=18$
$3x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=3x–7$.

Substitute into the first equation:
$2x+4(3x–7)=18$
$2x+12x–28=18$
$14x–28=18$
$14x=18+28$
$14x=46$
$x=\frac{46}{14}=\frac{23}{7}$.

Substitute $x=\frac{23}{7}$ into $y=3x–7$:
$y=3\left(\frac{23}{7}\right)–7$
$y=\frac{69}{7}–7$
$y=\frac{69}{7}–\frac{49}{7}$
$y=\frac{20}{7}$.

Thus, $x=\frac{23}{7}$ and $y=\frac{20}{7}$.

---

Example 75:
Solve the system of equations:
$7x–5y=15$
$4x+3y=13$

Solution using substitution:
From the second equation, solve for $y$:
$y=\frac{13–4x}{3}$.

Substitute into the first equation:
$7x–5\left(\frac{13–4x}{3}\right)=15$
$7x–\frac{5(13–4x)}{3}=15$
$7x–\frac{65–20x}{3}=15$.

Multiply by 3 to eliminate the denominator:
$3(7x–\frac{65–20x}{3})=3(15)$
$21x–65+20x=45$
$41x–65=45$
$41x=45+65$
$41x=110$
$x=\frac{110}{41}$.

Substitute $x=\frac{110}{41}$ into $y=\frac{13–4x}{3}$:
$y=\frac{13–4\left(\frac{110}{41}\right)}{3}$
$y=\frac{13–\frac{440}{41}}{3}$
$y=\frac{533–440}{123}$
$y=\frac{93}{123}=\frac{31}{41}$.

Thus, $x=\frac{110}{41}$ and $y=\frac{31}{41}$.

---

Example 76:
Solve the system of equations:
$3x+4y=19$
$2x–3y=6$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{19–3x}{4}$.

Substitute into the second equation:
$2x–3\left(\frac{19–3x}{4}\right)=6$
$2x–\frac{3(19–3x)}{4}=6$
$2x–\frac{57–9x}{4}=6$.

Multiply by 4 to eliminate the denominator:
$4(2x–\frac{57–9x}{4})=4(6)$
$8x–57+9x=24$
$17x–57=24$
$17x=24+57$
$17x=81$
$x=\frac{81}{17}$.

Substitute $x=\frac{81}{17}$ into $y=\frac{19–3x}{4}$:
$y=\frac{19–3\left(\frac{81}{17}\right)}{4}$
$y=\frac{19–\frac{243}{17}}{4}$
$y=\frac{323–243}{68}$
$y=\frac{80}{68}=\frac{40}{34}=\frac{20}{17}$.

Thus, $x=\frac{81}{17}$ and $y=\frac{20}{17}$.

---

Example 77:
Solve the system of equations:
$5x+y=12$
$3x–2y=7$

Solution using substitution:
From the first equation, solve for $y$:
$y=12–5x$.

Substitute into the second equation:
$3x–2(12–5x)=7$
$3x–24+10x=7$
$13x–24=7$
$13x=7+24$
$13x=31$
$x=\frac{31}{13}$.

Substitute $x=\frac{31}{13}$ into $y=12–5x$:
$y=12–5\left(\frac{31}{13}\right)$
$y=12–\frac{155}{13}$
$y=\frac{156}{13}–\frac{155}{13}$
$y=\frac{1}{13}$.

Thus, $x=\frac{31}{13}$ and $y=\frac{1}{13}$.

---

Example 78:
Solve the system of equations:
$2x+5y=17$
$4x–3y=6$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{17–2x}{5}$.

Substitute into the second equation:
$4x–3\left(\frac{17–2x}{5}\right)=6$
$4x–\frac{3(17–2x)}{5}=6$
$4x–\frac{51–6x}{5}=6$.

Multiply by 5 to eliminate the denominator:
$5(4x–\frac{51–6x}{5})=5(6)$
$20x–51+6x=30$
$26x–51=30$
$26x=30+51$
$26x=81$
$x=\frac{81}{26}$.

Substitute $x=\frac{81}{26}$ into $y=\frac{17–2x}{5}$:
$y=\frac{17–2\left(\frac{81}{26}\right)}{5}$
$y=\frac{17–\frac{162}{26}}{5}$
$y=\frac{442–162}{130}$
$y=\frac{280}{130}=\frac{14}{13}$.

Thus, $x=\frac{81}{26}$ and $y=\frac{14}{13}$.

---

Example 79:
Solve the system of equations:
$6x–y=9$
$3x+2y=8$

Solution using substitution:
From the first equation, solve for $y$:
$y=6x–9$.

Substitute into the second equation:
$3x+2(6x–9)=8$
$3x+12x–18=8$
$15x–18=8$
$15x=8+18$
$15x=26$
$x=\frac{26}{15}$.

Substitute $x=\frac{26}{15}$ into $y=6x–9$:
$y=6\left(\frac{26}{15}\right)–9$
$y=\frac{156}{15}–9$
$y=\frac{156}{15}–\frac{135}{15}$
$y=\frac{21}{15}=\frac{7}{5}$.

Thus, $x=\frac{26}{15}$ and $y=\frac{7}{5}$.

---

Example 80:
Solve the system of equations:
$4x+3y=22$
$5x–2y=13$

Solution using substitution:
From the second equation, solve for $y$:
$y=\frac{5x–13}{2}$.

Substitute into the first equation:
$4x+3\left(\frac{5x–13}{2}\right)=22$
$4x+\frac{3(5x–13)}{2}=22$
$4x+\frac{15x–39}{2}=22$.

Multiply by 2 to eliminate the denominator:
$2(4x+\frac{15x–39}{2})=2(22)$
$8x+15x–39=44$
$23x–39=44$
$23x=44+39$
$23x=83$
$x=\frac{83}{23}$.

Substitute $x=\frac{83}{23}$ into $y=\frac{5x–13}{2}$:
$y=\frac{5\left(\frac{83}{23}\right)–13}{2}$
$y=\frac{\frac{415}{23}–13}{2}$
$y=\frac{\frac{415}{23}–\frac{299}{23}}{2}$
$y=\frac{116}{23}\times \frac{1}{2}=\frac{58}{23}$.

Thus, $x=\frac{83}{23}$ and $y=\frac{58}{23}$.

---

Examples 81 to 90

---

Example 81:
Solve the system of equations:
$3x+5y=19$
$7x–2y=9$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{19–3x}{5}$.

Substitute into the second equation:
$7x–2\left(\frac{19–3x}{5}\right)=9$
$7x–\frac{2(19–3x)}{5}=9$
$7x–\frac{38–6x}{5}=9$.

Multiply by 5 to eliminate the denominator:
$5(7x–\frac{38–6x}{5})=5(9)$
$35x–38+6x=45$
$41x–38=45$
$41x=45+38$
$41x=83$
$x=\frac{83}{41}=2$.

Substitute $x=2$ into $y=\frac{19–3x}{5}$:
$y=\frac{19–3(2)}{5}$
$y=\frac{19–6}{5}$
$y=\frac{13}{5}$.

Thus, $x=2$ and $y=\frac{13}{5}$.

---

Example 82:
Solve the system of equations:
$4x–3y=8$
$5x+2y=14$

Solution using substitution:
From the second equation, solve for $y$:
$y=\frac{14–5x}{2}$.

Substitute into the first equation:
$4x–3\left(\frac{14–5x}{2}\right)=8$
$4x–\frac{3(14–5x)}{2}=8$
$4x–\frac{42–15x}{2}=8$.

Multiply by 2 to eliminate the denominator:
$2(4x–\frac{42–15x}{2})=2(8)$
$8x–42+15x=16$
$23x–42=16$
$23x=16+42$
$23x=58$
$x=\frac{58}{23}=\frac{29}{23}$.

Substitute $x=\frac{58}{23}$ into $y=\frac{14–5x}{2}$:
$y=\frac{14–5\left(\frac{58}{23}\right)}{2}$
$y=\frac{14–\frac{290}{23}}{2}$
$y=\frac{14}{5}$.

---

Example 83:
Solve the system of equations:
$3x+4y=12$
$5x–2y=10$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{12–3x}{4}$.

Substitute into the second equation:
$5x–2\left(\frac{12–3x}{4}\right)=10$
$5x–\frac{2(12–3x)}{4}=10$
$5x–\frac{24–6x}{4}=10$.

Multiply by 4 to eliminate the denominator:
$4(5x–\frac{24–6x}{4})=4(10)$
$20x–24+6x=40$
$26x–24=40$
$26x=40+24$
$26x=64$
$x=\frac{64}{26}=\frac{32}{13}$.

Substitute $x=\frac{32}{13}$ into $y=\frac{12–3x}{4}$:
$y=\frac{12–3\left(\frac{32}{13}\right)}{4}$
$y=\frac{12–\frac{96}{13}}{4}$
$y=\frac{156–96}{13}$
$y=\frac{60}{13}$.

Thus, $x=\frac{32}{13}$ and $y=\frac{60}{13}$.

---

Example 84:
Solve the system of equations:
$6x+y=15$
$4x–3y=11$

Solution using substitution:
From the first equation, solve for $y$:
$y=15–6x$.

Substitute into the second equation:
$4x–3(15–6x)=11$
$4x–45+18x=11$
$22x–45=11$
$22x=11+45$
$22x=56$
$x=\frac{56}{22}=\frac{28}{11}$.

Substitute $x=\frac{28}{11}$ into $y=15–6x$:
$y=15–6\left(\frac{28}{11}\right)$
$y=15–\frac{168}{11}$
$y=\frac{165}{11}–\frac{168}{11}$
$y=\frac{-3}{11}$.

Thus, $x=\frac{28}{11}$ and $y=\frac{-3}{11}$.

---

Example 85:
Solve the system of equations:
$3x–5y=17$
$2x+4y=10$

Solution using substitution:
From the second equation, solve for $y$:
$y=\frac{10–2x}{4}$.

Substitute into the first equation:
$3x–5\left(\frac{10–2x}{4}\right)=17$
$3x–\frac{5(10–2x)}{4}=17$
$3x–\frac{50–10x}{4}=17$.

Multiply by 4 to eliminate the denominator:
$4(3x–\frac{50–10x}{4})=4(17)$
$12x–50+10x=68$
$22x–50=68$
$22x=68+50$
$22x=118$
$x=\frac{118}{22}=\frac{59}{11}$.

Substitute $x=\frac{59}{11}$ into $y=\frac{10–2x}{4}$:
$y=\frac{10–2\left(\frac{59}{11}\right)}{4}$
$y=\frac{10–\frac{118}{11}}{4}$
$y=\frac{110–118}{44}$
$y=\frac{-8}{11}$.

Thus, $x=\frac{59}{11}$ and $y=\frac{-8}{11}$.

---

Example 86:
Solve the system of equations:
$7x+2y=25$
$5x–3y=14$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{25–7x}{2}$.

Substitute into the second equation:
$5x–3\left(\frac{25–7x}{2}\right)=14$
$5x–\frac{3(25–7x)}{2}=14$
$5x–\frac{75–21x}{2}=14$.

Multiply by 2 to eliminate the denominator:
$2(5x–\frac{75–21x}{2})=2(14)$
$10x–75+21x=28$
$31x–75=28$
$31x=28+75$
$31x=103$
$x=\frac{103}{31}$.

Substitute $x=\frac{103}{31}$ into $y=\frac{25–7x}{2}$:
$y=\frac{25–7\left(\frac{103}{31}\right)}{2}$
$y=\frac{25–\frac{721}{31}}{2}$
$y=\frac{-696}{62}$.

---

Example 87:
Solve the system of equations:
$2x+3y=14$
$4x–y=18$

Solution using substitution:
From the second equation, solve for $y$:
$y=4x–18$.

Substitute into the first equation:
$2x+3(4x–18)=14$
$2x+12x–54=14$
$14x–54=14$
$14x=14+54$
$14x=68$
$x=\frac{68}{14}=\frac{34}{7}$.

Substitute $x=\frac{34}{7}$ into $y=4x–18$:
$y=4\left(\frac{34}{7}\right)–18$
$y=\frac{136}{7}–18$
$y=\frac{136}{7}–\frac{126}{7}$
$y=\frac{10}{7}$.

Thus, $x=\frac{34}{7}$ and $y=\frac{10}{7}$.

---

Example 88:
Solve the system of equations:
$3x+4y=20$
$2x–5y=15$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{20–3x}{4}$.

Substitute into the second equation:
$2x–5\left(\frac{20–3x}{4}\right)=15$
$2x–\frac{5(20–3x)}{4}=15$
$2x–\frac{100–15x}{4}=15$.

Multiply by 4 to eliminate the denominator:
$4(2x–\frac{100–15x}{4})=4(15)$
$8x–100+15x=60$
$23x–100=60$
$23x=60+100$
$23x=160$
$x=\frac{160}{23}$.

Substitute $x=\frac{160}{23}$ into $y=\frac{20–3x}{4}$:
$y=\frac{20–3\left(\frac{160}{23}\right)}{4}$
$y=\frac{20–\frac{480}{23}}{4}$
$y=\frac{-260}{23}\times \frac{1}{4}=\frac{-65}{23}$.

Thus, $x=\frac{160}{23}$ and $y=\frac{-65}{23}$.

---

Example 89:
Solve the system of equations:
$4x+6y=30$
$5x–3y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=\frac{5x–7}{3}$.

Substitute into the first equation:
$4x+6\left(\frac{5x–7}{3}\right)=30$
$4x+\frac{6(5x–7)}{3}=30$
$4x+\frac{30x–42}{3}=30$.

Multiply by 3 to eliminate the denominator:
$3(4x+\frac{30x–42}{3})=3(30)$
$12x+30x–42=90$
$42x–42=90$
$42x=90+42$
$42x=132$
$x=\frac{132}{42}=\frac{22}{7}$.

Substitute $x=\frac{22}{7}$ into $y=\frac{5x–7}{3}$:
$y=\frac{5\left(\frac{22}{7}\right)–7}{3}$
$y=\frac{\frac{110}{7}–7}{3}$
$y=\frac{\frac{110}{7}–\frac{49}{7}}{3}$
$y=\frac{61}{21}$.

Thus, $x=\frac{22}{7}$ and $y=\frac{61}{21}$.

---

Example 90:
Solve the system of equations:
$6x–5y=11$
$3x+2y=7$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{6x–11}{5}$.

Substitute into the second equation:
$3x+2\left(\frac{6x–11}{5}\right)=7$
$3x+\frac{2(6x–11)}{5}=7$
$3x+\frac{12x–22}{5}=7$.

Multiply by 5 to eliminate the denominator:
$5(3x+\frac{12x–22}{5})=5(7)$
$15x+12x–22=35$
$27x–22=35$
$27x=35+22$
$27x=57$
$x=\frac{57}{27}=\frac{19}{9}$.

Substitute $x=\frac{19}{9}$ into $y=\frac{6x–11}{5}$:
$y=\frac{6\left(\frac{19}{9}\right)–11}{5}$
$y=\frac{\frac{114}{9}–11}{5}$
$y=\frac{\frac{114}{9}–\frac{99}{9}}{5}$
$y=\frac{15}{45}=\frac{1}{3}$.

Thus, $x=\frac{19}{9}$ and $y=\frac{1}{3}$.

---

Examples 91 to 100

---

Example 91:
Solve the system of equations:
$7x+4y=32$
$5x–y=13$

Solution using substitution:
From the second equation, solve for $y$:
$y=5x–13$.

Substitute into the first equation:
$7x+4(5x–13)=32$
$7x+20x–52=32$
$27x–52=32$
$27x=32+52$
$27x=84$
$x=\frac{84}{27}=\frac{28}{9}$.

Substitute $x=\frac{28}{9}$ into $y=5x–13$:
$y=5\left(\frac{28}{9}\right)–13$
$y=\frac{140}{9}–13$
$y=\frac{140}{9}–\frac{117}{9}$
$y=\frac{23}{9}$.

Thus, $x=\frac{28}{9}$ and $y=\frac{23}{9}$.

---

Example 92:
Solve the system of equations:
$4x+5y=18$
$3x–7y=4$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{18–4x}{5}$.

Substitute into the second equation:
$3x–7\left(\frac{18–4x}{5}\right)=4$
$3x–\frac{7(18–4x)}{5}=4$
$3x–\frac{126–28x}{5}=4$.

Multiply by 5 to eliminate the denominator:
$5(3x–\frac{126–28x}{5})=5(4)$
$15x–126+28x=20$
$43x–126=20$
$43x=20+126$
$43x=146$
$x=\frac{146}{43}$.

Substitute $x=\frac{146}{43}$ into $y=\frac{18–4x}{5}$:
$y=\frac{18–4\left(\frac{146}{43}\right)}{5}$
$y=\frac{18–\frac{584}{43}}{5}$
$y=\frac{774–584}{215}$
$y=\frac{190}{215}=\frac{38}{43}$.

Thus, $x=\frac{146}{43}$ and $y=\frac{38}{43}$.

---

Example 93:
Solve the system of equations:
$6x–5y=13$
$4x+y=11$

Solution using substitution:
From the second equation, solve for $y$:
$y=11–4x$.

Substitute into the first equation:
$6x–5(11–4x)=13$
$6x–55+20x=13$
$26x–55=13$
$26x=13+55$
$26x=68$
$x=\frac{68}{26}=\frac{34}{13}$.

Substitute $x=\frac{34}{13}$ into $y=11–4x$:
$y=11–4\left(\frac{34}{13}\right)$
$y=11–\frac{136}{13}$
$y=\frac{143}{13}–\frac{136}{13}$
$y=\frac{7}{13}$.

Thus, $x=\frac{34}{13}$ and $y=\frac{7}{13}$.

---

Example 94:
Solve the system of equations:
$3x+y=17$
$5x–3y=19$

Solution using substitution:
From the first equation, solve for $y$:
$y=17–3x$.

Substitute into the second equation:
$5x–3(17–3x)=19$
$5x–51+9x=19$
$14x–51=19$
$14x=19+51$
$14x=70$
$x=\frac{70}{14}=5$.

Substitute $x=5$ into $y=17–3x$:
$y=17–3(5)$
$y=17–15$
$y=2$.

Thus, $x=5$ and $y=2$.

---

Example 95:
Solve the system of equations:
$2x+3y=21$
$4x–2y=10$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{21–2x}{3}$.

Substitute into the second equation:
$4x–2\left(\frac{21–2x}{3}\right)=10$
$4x–\frac{2(21–2x)}{3}=10$
$4x–\frac{42–4x}{3}=10$.

Multiply by 3 to eliminate the denominator:
$3(4x–\frac{42–4x}{3})=3(10)$
$12x–42+4x=30$
$16x–42=30$
$16x=30+42$
$16x=72$
$x=\frac{72}{16}=4.5$.

Substitute $x=4.5$ into $y=\frac{21–2x}{3}$:
$y=\frac{21–2(4.5)}{3}$
$y=\frac{21–9}{3}$
$y=\frac{12}{3}=4$.

Thus, $x=4.5$ and $y=4$.

---

Example 96:
Solve the system of equations:
$7x+2y=23$
$5x–y=7$

Solution using substitution:
From the second equation, solve for $y$:
$y=5x–7$.

Substitute into the first equation:
$7x+2(5x–7)=23$
$7x+10x–14=23$
$17x–14=23$
$17x=23+14$
$17x=37$
$x=\frac{37}{17}$.

Substitute $x=\frac{37}{17}$ into $y=5x–7$:
$y=5\left(\frac{37}{17}\right)–7$
$y=\frac{185}{17}–7$
$y=\frac{185}{17}–\frac{119}{17}$
$y=\frac{66}{17}$.

Thus, $x=\frac{37}{17}$ and $y=\frac{66}{17}$.

---

Example 97:
Solve the system of equations:
$4x–3y=18$
$2x+5y=17$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{4x–18}{3}$.

Continuing from where we left off: $26x=141$
$x=\frac{141}{26}$
$x=\frac{141}{26}=\frac{141÷13}{26÷13}=\frac{10.85}{2}$.

Substitute $x=10.85$ into $y=\frac{4x–18}{3}$:
$y=\frac{4(10.85)–18}{3}$
$y=\frac{43.4–18}{3}$
$y=\frac{25.4}{3}=8.47$.

Thus, $x=10.85$ and $y=8.47$.

---

Example 98:
Solve the system of equations:
$5x+2y=12$
$3x–4y=9$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{12–5x}{2}$.

Substitute into the second equation:
$3x–4\left(\frac{12–5x}{2}\right)=9$
$3x–\frac{4(12–5x)}{2}=9$
$3x–\frac{48–20x}{2}=9$.

Multiply by 2 to eliminate the denominator:
$2(3x–\frac{48–20x}{2})=2(9)$
$6x–48+20x=18$
$26x–48=18$
$26x=18+48$
$26x=66$
$x=\frac{66}{26}=2.54$.

Substitute $x=2.54$ into $y=\frac{12–5x}{2}$:
$y=\frac{12–5(2.54)}{2}$
$y=\frac{12–12.7}{2}$
$y=\frac{-0.7}{2}=-0.35$.

Thus, $x=2.54$ and $y=-0.35$.

---

Example 99:
Solve the system of equations:
$2x+3y=10$
$4x–2y=8$

Solution using substitution:
From the first equation, solve for $y$:
$y=\frac{10–2x}{3}$.

Substitute into the second equation:
$4x–2\left(\frac{10–2x}{3}\right)=8$
$4x–\frac{2(10–2x)}{3}=8$
$4x–\frac{20–4x}{3}=8$.

Multiply by 3 to eliminate the denominator:
$3(4x–\frac{20–4x}{3})=3(8)$
$12x–20+4x=24$
$16x–20=24$
$16x=24+20$
$16x=44$
$x=\frac{44}{16}=2.75$.

Substitute $x=2.75$ into $y=\frac{10–2x}{3}$:
$y=\frac{10–2(2.75)}{3}$
$y=\frac{10–5.5}{3}$
$y=\frac{4.5}{3}=1.5$.

Thus, $x=2.75$ and $y=1.5$.

---

Example 100:
Solve the system of equations:
$x+y=7$
$2x–y=3$

Solution using substitution:
From the first equation, solve for $y$:
$y=7–x$.

Substitute into the second equation:
$2x–(7–x)=3$
$2x–7+x=3$
$3x–7=3$
$3x=3+7$
$3x=10$
$x=\frac{10}{3}=3.33$.

Substitute $x=3.33$ into $y=7–x$:
$y=7–3.33$
$y=3.67$.

Thus, $x=3.33$ and $y=3.67$.