Linear equations and inequalities form the foundation of algebra and are widely used in various real-world applications. A linear equation is an equation that makes a straight line when graphed. It contains variables raised to the power of 1, and its general form is $ax + b = c$. A linear inequality is similar to a linear equation but involves inequalities such as $>$, $<$, $ \geq $, or $ \leq $ instead of an equal sign.
This topic will cover the theory behind linear equations and inequalities, explain how to solve them, and provide 100 examples to solidify the concepts.
1. What is a Linear Equation?
A linear equation is any equation that can be written in the form $ ax + b = c $, where $ x $ is the variable, and $ a $, $ b $, and $ c $ are constants. The highest power of the variable is 1, which means the equation is linear.
For example:
- $ 3x + 5 = 11 $ is a linear equation.
- $ x – 4 = 8 $ is another linear equation.
Steps to Solve Linear Equations:
- Simplify: Simplify both sides of the equation if necessary by combining like terms.
- Isolate the Variable: Use inverse operations (addition, subtraction, multiplication, or division) to get the variable on one side of the equation.
- Solve for the Variable: Perform the required operations to find the value of the variable.
Example: Solve $ 3x + 5 = 11 $.
Solution:
$ 3x + 5 = 11 $
$ 3x = 11 – 5 $
$ 3x = 6 $
$ x = \frac{6}{3} $
$ x = 2 $
2. What is a Linear Inequality?
A linear inequality is similar to a linear equation but involves an inequality symbol ($<$, $>$, $ \leq $, or $ \geq $) instead of an equal sign. The general form is $ ax + b < c $, where $ x $ is the variable and $ a $, $ b $, and $ c $ are constants.
For example:
- $ 2x + 3 < 7 $ is a linear inequality.
- $ 5x – 1 \geq 9 $ is another linear inequality.
Steps to Solve Linear Inequalities:
- Simplify: Simplify both sides of the inequality.
- Isolate the Variable: Use inverse operations to move the variable to one side.
- Solve for the Variable: Perform the necessary operations.
- Flip the Sign: When multiplying or dividing both sides of an inequality by a negative number, remember to flip the inequality sign.
Example: Solve $ 2x + 3 < 7 $.
Solution:
$ 2x + 3 < 7 $
$ 2x < 7 – 3 $
$ 2x < 4 $
$ x < \frac{4}{2} $
$ x < 2 $
3. Solving Systems of Linear Equations
A system of linear equations consists of two or more linear equations with the same variables. The solution is the set of values that satisfy all the equations simultaneously.
Methods to Solve Systems of Equations:
- Substitution: Solve one equation for one variable and substitute it into the other equation.
- Elimination: Add or subtract equations to eliminate one variable, then solve for the remaining variable.
- Graphing: Graph both equations and find the point where they intersect.
Example: Solve the system:
$ x + y = 5 $
$ 2x – y = 3 $
Solution using substitution:
From the first equation:
$ y = 5 – x $
Substitute this into the second equation:
$ 2x – (5 – x) = 3 $
$ 2x – 5 + x = 3 $
$ 3x – 5 = 3 $
$ 3x = 3 + 5 $
$ 3x = 8 $
$ x = \frac{8}{3} $
Now, substitute $ x = \frac{8}{3} $ into $ y = 5 – x $:
$ y = 5 – \frac{8}{3} $
$ y = \frac{15}{3} – \frac{8}{3} $
$ y = \frac{7}{3} $
Thus, the solution is $ x = \frac{8}{3} $ and $ y = \frac{7}{3} $.
4. Graphing Linear Inequalities
Graphing linear inequalities involves shading a region on a graph that represents all the possible solutions. To graph a linear inequality:
- First, graph the boundary line (the equation obtained by replacing the inequality sign with an equal sign).
- Determine whether to use a solid line (for $ \leq $ or $ \geq $) or a dashed line (for $ < $ or $ > $).
- Test a point that is not on the boundary line to decide which side of the line to shade.
Example: Graph the inequality $ 3x – y \geq 4 $.
Solution:
- Graph the boundary line $ 3x – y = 4 $.
- Since the inequality symbol is $ \geq $, use a solid line.
- Test the point $ (0, 0) $:
$ 3(0) – (0) = 0 $, which is not greater than or equal to 4.
So, shade the region above the line.
5. 100 Examples of Linear Equations and Inequalities
Examples 1 to 10
Example 1: Solve the equation $ 4x + 2 = 10 $.
Solution:
$ 4x + 2 = 10 $
$ 4x = 10 – 2 $
$ 4x = 8 $
$ x = \frac{8}{4} $
$ x = 2 $
Example 2: Solve the inequality $ 3x – 5 < 10 $.
Solution:
$ 3x – 5 < 10 $
$ 3x < 10 + 5 $
$ 3x < 15 $
$ x < \frac{15}{3} $
$ x < 5 $
Example 3: Solve the system of equations:
$ 2x + y = 7 $
$ x – y = 1 $
Solution using substitution:
From the second equation:
$ y = x – 1 $
Substitute into the first equation:
$ 2x + (x – 1) = 7 $
$ 2x + x – 1 = 7 $
$ 3x – 1 = 7 $
$ 3x = 7 + 1 $
$ 3x = 8 $
$ x = \frac{8}{3} $
Now, substitute $ x = \frac{8}{3} $ into $ y = x – 1 $:
$ y = \frac{8}{3} – 1 $
$ y = \frac{8}{3} – \frac{3}{3} $
$ y = \frac{5}{3} $
Thus, the solution is $ x = \frac{8}{3} $ and $ y = \frac{5}{3} $.
Example 4: Solve the inequality $ 5x + 3 \leq 18 $.
Solution:
$ 5x + 3 \leq 18 $
$ 5x \leq 18 – 3 $
$ 5x \leq 15 $
$ x \leq \frac{15}{5} $
$ x \leq 3 $
Example 5: Solve $ 2x + 5 = 3x – 4 $.
Solution:
$ 2x + 5 = 3x – 4 $
$ 2x – 3x = -4 – 5 $
$ -x = -9 $
$ x = 9 $
Example 6: Solve the inequality $ 6x – 2 > 10 $.
Solution:
$ 6x – 2 > 10 $
$ 6x > 10 + 2 $
$ 6x > 12 $
$ x > \frac{12}{6} $
$ x > 2 $
Example 7: Solve the system of equations:
$ x + 2y = 8 $
$ 2x – y = 3 $
Solution using elimination:
From the second equation, multiply by 2: $ 2(x + 2y = 8) \rightarrow 2x + 4y = 16 $
Now, add the two equations: $ 2x – y = 3 $
$ 2x + 4y = 16 $
By adding these, we get: $ (2x + 4y) + (2x – y) = 16 + 3 $
$ 4x + 3y = 19 $.
Solve for $y$ by substitution.
Example 8: Solve the inequality $ 7x – 3 \geq 25 $.
Solution:
$ 7x – 3 \geq 25 $
$ 7x \geq 25 + 3 $
$ 7x \geq 28 $
$ x \geq \frac{28}{7} $
$ x \geq 4 $
Example 9: Solve $ 3x + 4 = 5x – 2 $.
Solution:
$ 3x + 4 = 5x – 2 $
$ 3x – 5x = -2 – 4 $
$ -2x = -6 $
$ x = \frac{-6}{-2} $
$ x = 3 $
Example 10: Solve the inequality $ 4x + 5 \leq 17 $.
Solution:
$ 4x + 5 \leq 17 $
$ 4x \leq 17 – 5 $
$ 4x \leq 12 $
$ x \leq \frac{12}{4} $
$ x \leq 3 $
Examples 11 to 20
Example 11: Solve the system of equations:
$ 2x + y = 6 $
$ 3x – 2y = 7 $
Solution using substitution:
From the first equation: $ y = 6 – 2x $.
Substitute into the second equation: $ 3x – 2(6 – 2x) = 7 $
$ 3x – 12 + 4x = 7 $
$ 7x – 12 = 7 $
$ 7x = 19 $
$ x = \frac{19}{7} $.
Now substitute $ x = \frac{19}{7} $ into $ y = 6 – 2x $: $ y = 6 – 2\left(\frac{19}{7}\right) $
$ y = \frac{42}{7} – \frac{38}{7} $
$ y = \frac{4}{7} $.
Thus, $ x = \frac{19}{7} $ and $ y = \frac{4}{7} $.
Example 12: Solve the inequality $ 6x – 9 < 15 $.
Solution:
$ 6x – 9 < 15 $
$ 6x < 15 + 9 $
$ 6x < 24 $
$ x < \frac{24}{6} $
$ x < 4 $
Example 13: Solve $ 5x – 7 = 2x + 8 $.
Solution:
$ 5x – 7 = 2x + 8 $
$ 5x – 2x = 8 + 7 $
$ 3x = 15 $
$ x = \frac{15}{3} $
$ x = 5 $
Example 14: Solve the inequality $ 4x + 7 \geq 23 $.
Solution:
$ 4x + 7 \geq 23 $
$ 4x \geq 23 – 7 $
$ 4x \geq 16 $
$ x \geq \frac{16}{4} $
$ x \geq 4 $
Example 15: Solve the system of equations:
$ x – y = 2 $
$ 2x + y = 7 $
Solution using substitution:
From the first equation: $ x = y + 2 $.
Substitute into the second equation: $ 2(y + 2) + y = 7 $
$ 2y + 4 + y = 7 $
$ 3y + 4 = 7 $
$ 3y = 7 – 4 $
$ 3y = 3 $
$ y = \frac{3}{3} $
$ y = 1 $.
Substitute $ y = 1 $ into $ x = y + 2 $: $ x = 1 + 2 = 3 $.
Thus, the solution is $ x = 3 $ and $ y = 1 $.
Example 16: Solve the inequality $ 2x + 5 \leq 11 $.
Solution:
$ 2x + 5 \leq 11 $
$ 2x \leq 11 – 5 $
$ 2x \leq 6 $
$ x \leq \frac{6}{2} $
$ x \leq 3 $
Example 17: Solve $ 7x – 3 = 4x + 6 $.
Solution:
$ 7x – 3 = 4x + 6 $
$ 7x – 4x = 6 + 3 $
$ 3x = 9 $
$ x = \frac{9}{3} $
$ x = 3 $
Example 18: Solve the inequality $ 5x + 4 < 20 $.
Solution:
$ 5x + 4 < 20 $
$ 5x < 20 – 4 $
$ 5x < 16 $
$ x < \frac{16}{5} $
$ x < 3.2 $
Example 19: Solve the system of equations:
$ 3x + y = 11 $
$ x – y = 2 $
Solution using substitution:
From the second equation: $ x = y + 2 $.
Substitute into the first equation: $ 3(y + 2) + y = 11 $
$ 3y + 6 + y = 11 $
$ 4y + 6 = 11 $
$ 4y = 11 – 6 $
$ 4y = 5 $
$ y = \frac{5}{4} $.
Now, substitute $ y = \frac{5}{4} $ into $ x = y + 2 $: $ x = \frac{5}{4} + 2 $
$ x = \frac{5}{4} + \frac{8}{4} $
$ x = \frac{13}{4} $.
Thus, $ x = \frac{13}{4} $ and $ y = \frac{5}{4} $.
Example 20: Solve the inequality $ 7x – 2 \geq 19 $.
Solution:
$ 7x – 2 \geq 19 $
$ 7x \geq 19 + 2 $
$ 7x \geq 21 $
$ x \geq \frac{21}{7} $
$ x \geq 3 $
Examples 21 to 30
Example 21: Solve the system of equations:
$ 2x + 3y = 12 $
$ x – y = 2 $
Solution using substitution:
From the second equation:
$ x = y + 2 $.
Substitute into the first equation:
$ 2(y + 2) + 3y = 12 $
$ 2y + 4 + 3y = 12 $
$ 5y + 4 = 12 $
$ 5y = 12 – 4 $
$ 5y = 8 $
$ y = \frac{8}{5} $.
Now substitute $ y = \frac{8}{5} $ into $ x = y + 2 $:
$ x = \frac{8}{5} + 2 $
$ x = \frac{8}{5} + \frac{10}{5} $
$ x = \frac{18}{5} $.
Thus, $ x = \frac{18}{5} $ and $ y = \frac{8}{5} $.
Example 22: Solve the inequality $ 5x – 3 < 12 $.
Solution:
$ 5x – 3 < 12 $
$ 5x < 12 + 3 $
$ 5x < 15 $
$ x < \frac{15}{5} $
$ x < 3 $
Example 23: Solve $ 4x + 6 = 2x + 12 $.
Solution:
$ 4x + 6 = 2x + 12 $
$ 4x – 2x = 12 – 6 $
$ 2x = 6 $
$ x = \frac{6}{2} $
$ x = 3 $
Example 24: Solve the inequality $ 3x + 7 \geq 16 $.
Solution:
$ 3x + 7 \geq 16 $
$ 3x \geq 16 – 7 $
$ 3x \geq 9 $
$ x \geq \frac{9}{3} $
$ x \geq 3 $
Example 25: Solve the system of equations:
$ x + 3y = 9 $
$ 2x – y = 7 $
Solution using substitution:
From the first equation:
$ x = 9 – 3y $.
Substitute into the second equation:
$ 2(9 – 3y) – y = 7 $
$ 18 – 6y – y = 7 $
$ 18 – 7y = 7 $
$ 7y = 18 – 7 $
$ 7y = 11 $
$ y = \frac{11}{7} $.
Substitute $ y = \frac{11}{7} $ into $ x = 9 – 3y $:
$ x = 9 – 3\left(\frac{11}{7}\right) $
$ x = \frac{63}{7} – \frac{33}{7} $
$ x = \frac{30}{7} $.
Thus, $ x = \frac{30}{7} $ and $ y = \frac{11}{7} $.
Example 26: Solve the inequality $ 4x – 5 \leq 11 $.
Solution:
$ 4x – 5 \leq 11 $
$ 4x \leq 11 + 5 $
$ 4x \leq 16 $
$ x \leq \frac{16}{4} $
$ x \leq 4 $
Example 27: Solve $ 6x – 7 = 4x + 9 $.
Solution:
$ 6x – 7 = 4x + 9 $
$ 6x – 4x = 9 + 7 $
$ 2x = 16 $
$ x = \frac{16}{2} $
$ x = 8 $
Example 28: Solve the inequality $ 3x + 2 < 14 $.
Solution:
$ 3x + 2 < 14 $
$ 3x < 14 – 2 $
$ 3x < 12 $
$ x < \frac{12}{3} $
$ x < 4 $
Example 29: Solve the system of equations:
$ 2x + 5y = 15 $
$ 3x – 4y = 11 $
Solution using elimination:
Multiply the first equation by 3 and the second equation by 2:
$ 3(2x + 5y = 15) \rightarrow 6x + 15y = 45 $
$ 2(3x – 4y = 11) \rightarrow 6x – 8y = 22 $.
Subtract the second equation from the first:
$ (6x + 15y) – (6x – 8y) = 45 – 22 $
$ 23y = 23 $
$ y = \frac{23}{23} $
$ y = 1 $.
Substitute $ y = 1 $ into $ 2x + 5y = 15 $:
$ 2x + 5(1) = 15 $
$ 2x + 5 = 15 $
$ 2x = 15 – 5 $
$ 2x = 10 $
$ x = \frac{10}{2} $
$ x = 5 $.
Thus, $ x = 5 $ and $ y = 1 $.
Example 30: Solve the inequality $ 8x – 3 \geq 21 $.
Solution:
$ 8x – 3 \geq 21 $
$ 8x \geq 21 + 3 $
$ 8x \geq 24 $
$ x \geq \frac{24}{8} $
$ x \geq 3 $
Examples 31 to 40
Example 31: Solve the system of equations:
$ x + 2y = 7 $
$ 2x – y = 4 $
Solution using substitution:
From the first equation:
$ x = 7 – 2y $.
Substitute into the second equation:
$ 2(7 – 2y) – y = 4 $
$ 14 – 4y – y = 4 $
$ 14 – 5y = 4 $
$ 5y = 14 – 4 $
$ 5y = 10 $
$ y = \frac{10}{5} $
$ y = 2 $.
Now, substitute $ y = 2 $ into $ x = 7 – 2y $:
$ x = 7 – 2(2) $
$ x = 7 – 4 $
$ x = 3 $.
Thus, $ x = 3 $ and $ y = 2 $.
Example 32: Solve the inequality $ 5x + 7 < 22 $.
Solution:
$ 5x + 7 < 22 $
$ 5x < 22 – 7 $
$ 5x < 15 $
$ x < \frac{15}{5} $
$ x < 3 $
Example 33: Solve the inequality $ 3x – 2 \geq 7 $.
Solution:
$ 3x – 2 \geq 7 $
$ 3x \geq 7 + 2 $
$ 3x \geq 9 $
$ x \geq \frac{9}{3} $
$ x \geq 3 $
Example 34: Solve the system of equations:
$ 4x + y = 12 $
$ 2x – y = 4 $
Solution using elimination:
Add both equations:
$ (4x + y) + (2x – y) = 12 + 4 $
$ 6x = 16 $
$ x = \frac{16}{6} = \frac{8}{3} $.
Substitute $ x = \frac{8}{3} $ into $ 4x + y = 12 $:
$ 4\left(\frac{8}{3}\right) + y = 12 $
$ \frac{32}{3} + y = 12 $
$ y = 12 – \frac{32}{3} $
$ y = \frac{36}{3} – \frac{32}{3} $
$ y = \frac{4}{3} $.
Thus, $ x = \frac{8}{3} $ and $ y = \frac{4}{3} $.
Example 35: Solve the inequality $ 7x + 4 < 25 $.
Solution:
$ 7x + 4 < 25 $
$ 7x < 25 – 4 $
$ 7x < 21 $
$ x < \frac{21}{7} $
$ x < 3 $
Example 36: Solve the system of equations:
$ 5x + 2y = 20 $
$ 3x – 4y = 10 $
Solution using substitution:
From the first equation:
$ 5x + 2y = 20 \rightarrow x = \frac{20 – 2y}{5} $.
Substitute into the second equation:
$ 3\left(\frac{20 – 2y}{5}\right) – 4y = 10 $
$ \frac{60 – 6y}{5} – 4y = 10 $
Multiply by 5 to eliminate the fraction:
$ 60 – 6y – 20y = 50 $
$ -26y = 50 – 60 $
$ -26y = -10 $
$ y = \frac{-10}{-26} = \frac{5}{13} $.
Substitute $ y = \frac{5}{13} $ into $ x = \frac{20 – 2y}{5} $:
$ x = \frac{20 – 2\left(\frac{5}{13}\right)}{5} $
$ x = \frac{20 – \frac{10}{13}}{5} $
$ x = \frac{\frac{260}{13} – \frac{10}{13}}{5} $
$ x = \frac{250}{13 \times 5} $
$ x = \frac{250}{65} $
$ x = \frac{50}{13} $.
Thus, $ x = \frac{50}{13} $ and $ y = \frac{5}{13} $.
Example 37: Solve the inequality $ 4x – 7 \geq 9 $.
Solution:
$ 4x – 7 \geq 9 $
$ 4x \geq 9 + 7 $
$ 4x \geq 16 $
$ x \geq \frac{16}{4} $
$ x \geq 4 $
Example 38: Solve the system of equations:
$ x + 3y = 6 $
$ 2x – y = 4 $
Solution using substitution:
From the first equation:
$ x = 6 – 3y $.
Substitute into the second equation:
$ 2(6 – 3y) – y = 4 $
$ 12 – 6y – y = 4 $
$ 12 – 7y = 4 $
$ 7y = 12 – 4 $
$ 7y = 8 $
$ y = \frac{8}{7} $.
Substitute $ y = \frac{8}{7} $ into $ x = 6 – 3y $:
$ x = 6 – 3\left(\frac{8}{7}\right) $
$ x = \frac{42}{7} – \frac{24}{7} $
$ x = \frac{18}{7} $.
Thus, $ x = \frac{18}{7} $ and $ y = \frac{8}{7} $.
Example 39: Solve the inequality $ 2x + 5 < 18 $.
Solution:
$ 2x + 5 < 18 $
$ 2x < 18 – 5 $
$ 2x < 13 $
$ x < \frac{13}{2} $
$ x < 6.5 $
Example 40: Solve the system of equations:
$ 3x + 4y = 24 $
$ 2x – 3y = 6 $
Solution using elimination:
Multiply the first equation by 3 and the second equation by 4:
$ 3(3x + 4y = 24) \rightarrow 9x + 12y = 72 $
$ 4(2x – 3y = 6) \rightarrow 8x – 12y = 24 $.
Add both equations:
$ (9x + 12y) + (8x – 12y) = 72 + 24 $
$ 17x = 96 $
$ x = \frac{96}{17} $
$ x = \frac{96}{17} $.
Now substitute $ x = \frac{96}{17} $ into $ 3x + 4y = 24 $:
$ 3\left(\frac{96}{17}\right) + 4y = 24 $
$ \frac{288}{17} + 4y = 24 $
$ 4y = 24 – \frac{288}{17} $
$ 4y = \frac{408}{17} – \frac{288}{17} $
$ 4y = \frac{120}{17} $
$ y = \frac{120}{17 \times 4} $
$ y = \frac{30}{17} $.
Thus, $ x = \frac{96}{17} $ and $ y = \frac{30}{17} $.
Examples 41 to 50
Example 41: Solve the system of equations:
$ x – 2y = 4 $
$ 3x + y = 11 $
Solution using substitution:
From the first equation:
$ x = 4 + 2y $.
Substitute into the second equation:
$ 3(4 + 2y) + y = 11 $
$ 12 + 6y + y = 11 $
$ 12 + 7y = 11 $
$ 7y = 11 – 12 $
$ 7y = -1 $
$ y = \frac{-1}{7} $.
Substitute $ y = \frac{-1}{7} $ into $ x = 4 + 2y $:
$ x = 4 + 2\left(\frac{-1}{7}\right) $
$ x = 4 – \frac{2}{7} $
$ x = \frac{28}{7} – \frac{2}{7} $
$ x = \frac{26}{7} $.
Thus, $ x = \frac{26}{7} $ and $ y = \frac{-1}{7} $.
Example 42: Solve the inequality $ 5x – 4 \geq 16 $.
Solution:
$ 5x – 4 \geq 16 $
$ 5x \geq 16 + 4 $
$ 5x \geq 20 $
$ x \geq \frac{20}{5} $
$ x \geq 4 $
Example 43: Solve the system of equations:
$ 4x + 3y = 21 $
$ x – y = 2 $
Solution using elimination:
Multiply the second equation by 3:
$ 3(x – y = 2) \rightarrow 3x – 3y = 6 $.
Now add the two equations:
$ 4x + 3y + 3x – 3y = 21 + 6 $
$ 7x = 27 $
$ x = \frac{27}{7} $.
Substitute $ x = \frac{27}{7} $ into $ x – y = 2 $:
$ \frac{27}{7} – y = 2 $
$ y = \frac{27}{7} – 2 $
$ y = \frac{27}{7} – \frac{14}{7} $
$ y = \frac{13}{7} $.
Thus, $ x = \frac{27}{7} $ and $ y = \frac{13}{7} $.
Example 44: Solve the inequality $ 3x + 7 < 19 $.
Solution:
$ 3x + 7 < 19 $
$ 3x < 19 – 7 $
$ 3x < 12 $
$ x < \frac{12}{3} $
$ x < 4 $
Example 45: Solve the system of equations:
$ 2x + y = 7 $
$ 5x – 2y = 11 $
Solution using substitution:
From the first equation:
$ y = 7 – 2x $.
Substitute into the second equation:
$ 5x – 2(7 – 2x) = 11 $
$ 5x – 14 + 4x = 11 $
$ 9x – 14 = 11 $
$ 9x = 11 + 14 $
$ 9x = 25 $
$ x = \frac{25}{9} $.
Substitute $ x = \frac{25}{9} $ into $ y = 7 – 2x $:
$ y = 7 – 2\left(\frac{25}{9}\right) $
$ y = 7 – \frac{50}{9} $
$ y = \frac{63}{9} – \frac{50}{9} $
$ y = \frac{13}{9} $.
Thus, $ x = \frac{25}{9} $ and $ y = \frac{13}{9} $.
Example 46: Solve the inequality $ 6x – 3 \geq 9 $.
Solution:
$ 6x – 3 \geq 9 $
$ 6x \geq 9 + 3 $
$ 6x \geq 12 $
$ x \geq \frac{12}{6} $
$ x \geq 2 $
Example 47: Solve the system of equations:
$ 3x + y = 9 $
$ 2x – y = 4 $
Solution using elimination:
Add both equations:
$ (3x + y) + (2x – y) = 9 + 4 $
$ 5x = 13 $
$ x = \frac{13}{5} $.
Substitute $ x = \frac{13}{5} $ into $ 3x + y = 9 $:
$ 3\left(\frac{13}{5}\right) + y = 9 $
$ \frac{39}{5} + y = 9 $
$ 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 48: Solve the inequality $ 8x + 4 < 28 $.
Solution:
$ 8x + 4 < 28 $
$ 8x < 28 – 4 $
$ 8x < 24 $
$ x < \frac{24}{8} $
$ x < 3 $
Example 49: Solve the system of equations:
$ x + 4y = 14 $
$ 3x – 2y = 12 $
Solution using substitution:
From the first equation:
$ x = 14 – 4y $.
Substitute into the second equation:
$ 3(14 – 4y) – 2y = 12 $
$ 42 – 12y – 2y = 12 $
$ 42 – 14y = 12 $
$ -14y = 12 – 42 $
$ -14y = -30 $
$ y = \frac{-30}{-14} = \frac{15}{7} $.
Substitute $ y = \frac{15}{7} $ into $ x = 14 – 4y $:
$ x = 14 – 4\left(\frac{15}{7}\right) $
$ x = \frac{98}{7} – \frac{60}{7} $
$ x = \frac{38}{7} $.
Thus, $ x = \frac{38}{7} $ and $ y = \frac{15}{7} $.
Example 50: Solve the inequality $ 7x – 5 \geq 16 $.
Solution:
$ 7x – 5 \geq 16 $
$ 7x \geq 16 + 5 $
$ 7x \geq 21 $
$ x \geq \frac{21}{7} $
$ x \geq 3 $
Examples 51 to 60
Example 51: Solve the system of equations:
$ 2x + 3y = 13 $
$ x – y = 2 $
Solution using substitution:
From the second equation:
$ x = y + 2 $.
Substitute into the first equation:
$ 2(y + 2) + 3y = 13 $
$ 2y + 4 + 3y = 13 $
$ 5y + 4 = 13 $
$ 5y = 13 – 4 $
$ 5y = 9 $
$ y = \frac{9}{5} $.
Substitute $ y = \frac{9}{5} $ into $ x = y + 2 $:
$ x = \frac{9}{5} + 2 $
$ x = \frac{9}{5} + \frac{10}{5} $
$ x = \frac{19}{5} $.
Thus, $ x = \frac{19}{5} $ and $ y = \frac{9}{5} $.
Example 52: Solve the inequality $ 4x + 6 \leq 18 $.
Solution:
$ 4x + 6 \leq 18 $
$ 4x \leq 18 – 6 $
$ 4x \leq 12 $
$ x \leq \frac{12}{4} $
$ x \leq 3 $
Example 53: Solve the system of equations:
$ 5x + y = 20 $
$ 2x – y = 4 $
Solution using elimination:
Add both equations:
$ (5x + y) + (2x – y) = 20 + 4 $
$ 7x = 24 $
$ x = \frac{24}{7} $.
Substitute $ x = \frac{24}{7} $ into $ 5x + y = 20 $:
$ 5\left(\frac{24}{7}\right) + y = 20 $
$ \frac{120}{7} + y = 20 $
$ y = 20 – \frac{120}{7} $
$ y = \frac{140}{7} – \frac{120}{7} $
$ y = \frac{20}{7} $.
Thus, $ x = \frac{24}{7} $ and $ y = \frac{20}{7} $.
Example 54: Solve the inequality $ 7x – 9 < 12 $.
Solution:
$ 7x – 9 < 12 $
$ 7x < 12 + 9 $
$ 7x < 21 $
$ x < \frac{21}{7} $
$ x < 3 $
Example 55: Solve the system of equations:
$ 3x + y = 10 $
$ 4x – y = 12 $
Solution using elimination:
Add both equations:
$ (3x + y) + (4x – y) = 10 + 12 $
$ 7x = 22 $
$ x = \frac{22}{7} $.
Substitute $ x = \frac{22}{7} $ into $ 3x + y = 10 $:
$ 3\left(\frac{22}{7}\right) + y = 10 $
$ \frac{66}{7} + y = 10 $
$ y = 10 – \frac{66}{7} $
$ y = \frac{70}{7} – \frac{66}{7} $
$ y = \frac{4}{7} $.
Thus, $ x = \frac{22}{7} $ and $ y = \frac{4}{7} $.
Example 56: Solve the inequality $ 8x + 4 \geq 28 $.
Solution:
$ 8x + 4 \geq 28 $
$ 8x \geq 28 – 4 $
$ 8x \geq 24 $
$ x \geq \frac{24}{8} $
$ x \geq 3 $
Example 57: Solve the system of equations:
$ 6x + 2y = 18 $
$ 4x – 3y = 10 $
Solution using substitution:
From the first equation:
$ 6x + 2y = 18 \rightarrow x = \frac{18 – 2y}{6} $.
Substitute into the second equation:
$ 4\left(\frac{18 – 2y}{6}\right) – 3y = 10 $
$ \frac{72 – 8y}{6} – 3y = 10 $
Multiply by 6 to eliminate the fraction:
$ 72 – 8y – 18y = 60 $
$ 72 – 26y = 60 $
$ -26y = 60 – 72 $
$ -26y = -12 $
$ y = \frac{-12}{-26} = \frac{6}{13} $.
Substitute $ y = \frac{6}{13} $ into $ x = \frac{18 – 2y}{6} $:
$ x = \frac{18 – 2\left(\frac{6}{13}\right)}{6} $
$ x = \frac{18 – \frac{12}{13}}{6} $
$ x = \frac{\frac{234}{13} – \frac{12}{13}}{6} $
$ x = \frac{222}{13 \times 6} $
$ x = \frac{111}{39} $.
Thus, $ x = \frac{111}{39} $ and $ y = \frac{6}{13} $.
Example 58: Solve the inequality $ 5x – 7 < 18 $.
Solution:
$ 5x – 7 < 18 $
$ 5x < 18 + 7 $
$ 5x < 25 $
$ x < \frac{25}{5} $
$ x < 5 $
Example 59: Solve the system of equations:
$ x + 3y = 11 $
$ 2x – y = 4 $
Solution using substitution:
From the first equation:
$ x = 11 – 3y $.
Substitute into the second equation:
$ 2(11 – 3y) – y = 4 $
$ 22 – 6y – y = 4 $
$ 22 – 7y = 4 $
$ 7y = 22 – 4 $
$ 7y = 18 $
$ y = \frac{18}{7} $.
Substitute $ y = \frac{18}{7} $ into $ x = 11 – 3y $:
$ x = 11 – 3\left(\frac{18}{7}\right) $
$ x = \frac{77}{7} – \frac{54}{7} $
$ x = \frac{23}{7} $.
Thus, $ x = \frac{23}{7} $ and $ y = \frac{18}{7} $.
Example 60: Solve the inequality $ 4x + 3 \geq 15 $.
Solution:
$ 4x + 3 \geq 15 $
$ 4x \geq 15 – 3 $
$ 4x \geq 12 $
$ x \geq \frac{12}{4} $
$ x \geq 3 $
Examples 61 to 70
Example 61: Solve the system of equations:
$ 2x + 5y = 17 $
$ 3x – 2y = 4 $
Solution using substitution:
From the first equation:
$ x = \frac{17 – 5y}{2} $.
Substitute into the second equation:
$ 3\left(\frac{17 – 5y}{2}\right) – 2y = 4 $
$ \frac{51 – 15y}{2} – 2y = 4 $
Multiply by 2 to eliminate the fraction:
$ 51 – 15y – 4y = 8 $
$ 51 – 19y = 8 $
$ -19y = 8 – 51 $
$ -19y = -43 $
$ y = \frac{-43}{-19} = \frac{43}{19} $.
Substitute $ y = \frac{43}{19} $ into $ x = \frac{17 – 5y}{2} $:
$ x = \frac{17 – 5\left(\frac{43}{19}\right)}{2} $
$ x = \frac{17 – \frac{215}{19}}{2} $
$ x = \frac{\frac{323}{19} – \frac{215}{19}}{2} $
$ x = \frac{108}{19 \times 2} $
$ x = \frac{108}{38} $
$ x = \frac{54}{19} $.
Thus, $ x = \frac{54}{19} $ and $ y = \frac{43}{19} $.
Example 62: Solve the inequality $ 6x + 8 \leq 26 $.
Solution:
$ 6x + 8 \leq 26 $
$ 6x \leq 26 – 8 $
$ 6x \leq 18 $
$ x \leq \frac{18}{6} $
$ x \leq 3 $
Example 63: Solve the system of equations:
$ 4x + y = 13 $
$ 2x – y = 3 $
Solution using elimination:
Add both equations:
$ (4x + y) + (2x – y) = 13 + 3 $
$ 6x = 16 $
$ x = \frac{16}{6} = \frac{8}{3} $.
Substitute $ x = \frac{8}{3} $ into $ 4x + y = 13 $:
$ 4\left(\frac{8}{3}\right) + y = 13 $
$ \frac{32}{3} + y = 13 $
$ y = 13 – \frac{32}{3} $
$ y = \frac{39}{3} – \frac{32}{3} $
$ y = \frac{7}{3} $.
Thus, $ x = \frac{8}{3} $ and $ y = \frac{7}{3} $.
Example 64: Solve the inequality $ 4x – 9 \geq 3 $.
Solution:
$ 4x – 9 \geq 3 $
$ 4x \geq 3 + 9 $
$ 4x \geq 12 $
$ x \geq \frac{12}{4} $
$ x \geq 3 $
Example 65: Solve the system of equations:
$ 5x + y = 14 $
$ 3x – 2y = 7 $
Solution using substitution:
From the first equation:
$ y = 14 – 5x $.
Substitute into the second equation:
$ 3x – 2(14 – 5x) = 7 $
$ 3x – 28 + 10x = 7 $
$ 13x – 28 = 7 $
$ 13x = 7 + 28 $
$ 13x = 35 $
$ x = \frac{35}{13} $.
Substitute $ x = \frac{35}{13} $ into $ y = 14 – 5x $:
$ y = 14 – 5\left(\frac{35}{13}\right) $
$ y = \frac{182}{13} – \frac{175}{13} $
$ y = \frac{7}{13} $.
Thus, $ x = \frac{35}{13} $ and $ y = \frac{7}{13} $.
Example 66: Solve the inequality $ 5x – 6 < 9 $.
Solution:
$ 5x – 6 < 9 $
$ 5x < 9 + 6 $
$ 5x < 15 $
$ x < \frac{15}{5} $
$ x < 3 $
Example 67: Solve the system of equations:
$ 2x + 3y = 9 $
$ 4x – y = 7 $
Solution using substitution:
From the second equation:
$ y = 4x – 7 $.
Substitute into the first equation:
$ 2x + 3(4x – 7) = 9 $
$ 2x + 12x – 21 = 9 $
$ 14x – 21 = 9 $
$ 14x = 9 + 21 $
$ 14x = 30 $
$ x = \frac{30}{14} = \frac{15}{7} $.
Substitute $ x = \frac{15}{7} $ into $ y = 4x – 7 $:
$ y = 4\left(\frac{15}{7}\right) – 7 $
$ y = \frac{60}{7} – 7 $
$ y = \frac{60}{7} – \frac{49}{7} $
$ y = \frac{11}{7} $.
Thus, $ x = \frac{15}{7} $ and $ y = \frac{11}{7} $.
Example 68: Solve the inequality $ 3x + 5 \leq 11 $.
Solution:
$ 3x + 5 \leq 11 $
$ 3x \leq 11 – 5 $
$ 3x \leq 6 $
$ x \leq \frac{6}{3} $
$ x \leq 2 $
Example 69: Solve the system of equations:
$ 3x + 2y = 10 $
$ x – y = 2 $
Solution using substitution:
From the second equation:
$ y = x – 2 $.
Substitute into the first equation:
$ 3x + 2(x – 2) = 10 $
$ 3x + 2x – 4 = 10 $
$ 5x – 4 = 10 $
$ 5x = 10 + 4 $
$ 5x = 14 $
$ x = \frac{14}{5} $.
Substitute $ x = \frac{14}{5} $ into $ y = x – 2 $:
$ y = \frac{14}{5} – 2 $
$ y = \frac{14}{5} – \frac{10}{5} $
$ y = \frac{4}{5} $.
Thus, $ x = \frac{14}{5} $ and $ y = \frac{4}{5} $.
Example 70: Solve the inequality $ 6x – 8 < 16 $.
Solution:
$ 6x – 8 < 16 $
$ 6x < 16 + 8 $
$ 6x < 24 $
$ x < \frac{24}{6} $
$ x < 4 $
Examples 71 to 80
Example 71: Solve the system of equations:
$ 5x + 2y = 14 $
$ 4x – y = 5 $
Solution using substitution:
From the second equation:
$ y = 4x – 5 $.
Substitute into the first equation:
$ 5x + 2(4x – 5) = 14 $
$ 5x + 8x – 10 = 14 $
$ 13x – 10 = 14 $
$ 13x = 14 + 10 $
$ 13x = 24 $
$ x = \frac{24}{13} $.
Substitute $ x = \frac{24}{13} $ into $ y = 4x – 5 $:
$ y = 4\left(\frac{24}{13}\right) – 5 $
$ y = \frac{96}{13} – 5 $
$ y = \frac{96}{13} – \frac{65}{13} $
$ y = \frac{31}{13} $.
Thus, $ x = \frac{24}{13} $ and $ y = \frac{31}{13} $.
Example 72: Solve the inequality $ 8x – 9 \geq 15 $.
Solution:
$ 8x – 9 \geq 15 $
$ 8x \geq 15 + 9 $
$ 8x \geq 24 $
$ x \geq \frac{24}{8} $
$ x \geq 3 $
Example 73: Solve the system of equations:
$ 3x + 5y = 19 $
$ 2x – y = 6 $
Solution using elimination:
Multiply the second equation by 5:
$ 5(2x – y = 6) \rightarrow 10x – 5y = 30 $.
Now add both equations:
$ (3x + 5y) + (10x – 5y) = 19 + 30 $
$ 13x = 49 $
$ x = \frac{49}{13} $.
Substitute $ x = \frac{49}{13} $ into $ 2x – y = 6 $:
$ 2\left(\frac{49}{13}\right) – y = 6 $
$ \frac{98}{13} – y = 6 $
$ y = \frac{98}{13} – 6 $
$ y = \frac{98}{13} – \frac{78}{13} $
$ y = \frac{20}{13} $.
Thus, $ x = \frac{49}{13} $ and $ y = \frac{20}{13} $.
Example 74: Solve the inequality $ 7x + 4 < 23 $.
Solution:
$ 7x + 4 < 23 $
$ 7x < 23 – 4 $
$ 7x < 19 $
$ x < \frac{19}{7} $
$ x < 2.71 $
Example 75: Solve the system of equations:
$ 2x + 4y = 12 $
$ 3x – y = 5 $
Solution using substitution:
From the first equation:
$ x = \frac{12 – 4y}{2} $.
Substitute into the second equation:
$ 3\left(\frac{12 – 4y}{2}\right) – y = 5 $
$ \frac{36 – 12y}{2} – y = 5 $
Multiply by 2 to eliminate the fraction:
$ 36 – 12y – 2y = 10 $
$ 36 – 14y = 10 $
$ -14y = 10 – 36 $
$ -14y = -26 $
$ y = \frac{-26}{-14} = \frac{13}{7} $.
Substitute $ y = \frac{13}{7} $ into $ x = \frac{12 – 4y}{2} $:
$ x = \frac{12 – 4\left(\frac{13}{7}\right)}{2} $
$ x = \frac{12 – \frac{52}{7}}{2} $
$ x = \frac{\frac{84}{7} – \frac{52}{7}}{2} $
$ x = \frac{32}{7 \times 2} $
$ x = \frac{32}{14} = \frac{16}{7} $.
Thus, $ x = \frac{16}{7} $ and $ y = \frac{13}{7} $.
Example 76: Solve the inequality $ 5x – 8 \leq 7 $.
Solution:
$ 5x – 8 \leq 7 $
$ 5x \leq 7 + 8 $
$ 5x \leq 15 $
$ x \leq \frac{15}{5} $
$ x \leq 3 $
Example 77: Solve the system of equations:
$ 4x + y = 11 $
$ 2x – y = 3 $
Solution using elimination:
Add both equations:
$ (4x + y) + (2x – y) = 11 + 3 $
$ 6x = 14 $
$ x = \frac{14}{6} = \frac{7}{3} $.
Substitute $ x = \frac{7}{3} $ into $ 4x + y = 11 $:
$ 4\left(\frac{7}{3}\right) + y = 11 $
$ \frac{28}{3} + y = 11 $
$ y = 11 – \frac{28}{3} $
$ y = \frac{33}{3} – \frac{28}{3} $
$ y = \frac{5}{3} $.
Thus, $ x = \frac{7}{3} $ and $ y = \frac{5}{3} $.
Example 78: Solve the inequality $ 4x + 9 \geq 21 $.
Solution:
$ 4x + 9 \geq 21 $
$ 4x \geq 21 – 9 $
$ 4x \geq 12 $
$ x \geq \frac{12}{4} $
$ x \geq 3 $
Example 79: Solve the system of equations:
$ x + 2y = 5 $
$ 3x – y = 8 $
Solution using substitution:
From the first equation:
$ x = 5 – 2y $.
Substitute into the second equation:
$ 3(5 – 2y) – y = 8 $
$ 15 – 6y – y = 8 $
$ 15 – 7y = 8 $
$ 7y = 15 – 8 $
$ 7y = 7 $
$ y = \frac{7}{7} $
$ y = 1 $.
Substitute $ y = 1 $ into $ x = 5 – 2y $:
$ x = 5 – 2(1) $
$ x = 5 – 2 $
$ x = 3 $.
Thus, $ x = 3 $ and $ y = 1 $.
Example 80: Solve the inequality $ 6x – 4 < 14 $.
Solution:
$ 6x – 4 < 14 $
$ 6x < 14 + 4 $
$ 6x < 18 $
$ x < \frac{18}{6} $
$ x < 3 $
Examples 81 to 90
Example 81: Solve the system of equations:
$ 2x + 3y = 15 $
$ x – 4y = -1 $
Solution using substitution:
From the second equation:
$ x = 4y – 1 $.
Substitute into the first equation:
$ 2(4y – 1) + 3y = 15 $
$ 8y – 2 + 3y = 15 $
$ 11y – 2 = 15 $
$ 11y = 15 + 2 $
$ 11y = 17 $
$ y = \frac{17}{11} $.
Substitute $ y = \frac{17}{11} $ into $ x = 4y – 1 $:
$ x = 4\left(\frac{17}{11}\right) – 1 $
$ x = \frac{68}{11} – 1 $
$ x = \frac{68}{11} – \frac{11}{11} $
$ x = \frac{57}{11} $.
Thus, $ x = \frac{57}{11} $ and $ y = \frac{17}{11} $.
Example 82: Solve the inequality $ 3x + 4 \leq 19 $.
Solution:
$ 3x + 4 \leq 19 $
$ 3x \leq 19 – 4 $
$ 3x \leq 15 $
$ x \leq \frac{15}{3} $
$ x \leq 5 $
Example 83: Solve the system of equations:
$ 5x + 2y = 18 $
$ 3x – y = 7 $
Solution using elimination:
Multiply the second equation by 2:
$ 2(3x – y = 7) \rightarrow 6x – 2y = 14 $.
Now add both equations:
$ (5x + 2y) + (6x – 2y) = 18 + 14 $
$ 11x = 32 $
$ x = \frac{32}{11} $.
Substitute $ x = \frac{32}{11} $ into $ 3x – y = 7 $:
$ 3\left(\frac{32}{11}\right) – y = 7 $
$ \frac{96}{11} – y = 7 $
$ y = \frac{96}{11} – 7 $
$ y = \frac{96}{11} – \frac{77}{11} $
$ y = \frac{19}{11} $.
Thus, $ x = \frac{32}{11} $ and $ y = \frac{19}{11} $.
Example 84: Solve the inequality $ 5x – 3 < 12 $.
Solution:
$ 5x – 3 < 12 $
$ 5x < 12 + 3 $
$ 5x < 15 $
$ x < \frac{15}{5} $
$ x < 3 $
Example 85: Solve the system of equations:
$ 4x + y = 19 $
$ 2x – y = 1 $
Solution using elimination:
Add both equations:
$ (4x + y) + (2x – y) = 19 + 1 $
$ 6x = 20 $
$ x = \frac{20}{6} = \frac{10}{3} $.
Substitute $ x = \frac{10}{3} $ into $ 4x + y = 19 $:
$ 4\left(\frac{10}{3}\right) + y = 19 $
$ \frac{40}{3} + y = 19 $
$ y = 19 – \frac{40}{3} $
$ y = \frac{57}{3} – \frac{40}{3} $
$ y = \frac{17}{3} $.
Thus, $ x = \frac{10}{3} $ and $ y = \frac{17}{3} $.
Example 86: Solve the inequality $ 6x – 7 \geq 11 $.
Solution:
$ 6x – 7 \geq 11 $
$ 6x \geq 11 + 7 $
$ 6x \geq 18 $
$ x \geq \frac{18}{6} $
$ x \geq 3 $
Example 87: Solve the system of equations:
$ x + 2y = 8 $
$ 2x – y = 3 $
Solution using substitution:
From the first equation:
$ x = 8 – 2y $.
Substitute into the second equation:
$ 2(8 – 2y) – y = 3 $
$ 16 – 4y – y = 3 $
$ 16 – 5y = 3 $
$ 5y = 16 – 3 $
$ 5y = 13 $
$ y = \frac{13}{5} $.
Substitute $ y = \frac{13}{5} $ into $ x = 8 – 2y $:
$ x = 8 – 2\left(\frac{13}{5}\right) $
$ x = 8 – \frac{26}{5} $
$ x = \frac{40}{5} – \frac{26}{5} $
$ x = \frac{14}{5} $.
Thus, $ x = \frac{14}{5} $ and $ y = \frac{13}{5} $.
Example 88: Solve the inequality $ 4x + 7 \leq 21 $.
Solution:
$ 4x + 7 \leq 21 $
$ 4x \leq 21 – 7 $
$ 4x \leq 14 $
$ x \leq \frac{14}{4} $
$ x \leq 3.5 $
Example 89: Solve the system of equations:
$ 2x + y = 10 $
$ 5x – 3y = 1 $
Solution using substitution:
From the first equation:
$ y = 10 – 2x $.
Substitute into the second equation:
$ 5x – 3(10 – 2x) = 1 $
$ 5x – 30 + 6x = 1 $
$ 11x – 30 = 1 $
$ 11x = 1 + 30 $
$ 11x = 31 $
$ x = \frac{31}{11} $.
Substitute $ x = \frac{31}{11} $ into $ y = 10 – 2x $:
$ y = 10 – 2\left(\frac{31}{11}\right) $
$ y = 10 – \frac{62}{11} $
$ y = \frac{110}{11} – \frac{62}{11} $
$ y = \frac{48}{11} $.
Thus, $ x = \frac{31}{11} $ and $ y = \frac{48}{11} $.
Example 90: Solve the inequality $ 3x – 8 < 7 $.
Solution:
$ 3x – 8 < 7 $
$ 3x < 7 + 8 $
$ 3x < 15 $
$ x < \frac{15}{3} $
$ x < 5 $
Linear Equations and Inequalities
Linear equations and inequalities form the foundation of algebra and are widely used in various real-world applications. A linear equation is an equation that makes a straight line when graphed. It contains variables raised to the power of 1, and its general form is $ax + b = c$. A linear inequality is similar to a linear equation but involves inequalities such as $>$, $<$, $ \geq $, or $ \leq $ instead of an equal sign.
This topic will cover the theory behind linear equations and inequalities, explain how to solve them, and provide 100 examples to solidify the concepts.
1. What is a Linear Equation?
A linear equation is any equation that can be written in the form $ ax + b = c $, where $ x $ is the variable, and $ a $, $ b $, and $ c $ are constants. The highest power of the variable is 1, which means the equation is linear.
For example:
- $ 3x + 5 = 11 $ is a linear equation.
- $ x – 4 = 8 $ is another linear equation.
Steps to Solve Linear Equations:
- Simplify: Simplify both sides of the equation if necessary by combining like terms.
- Isolate the Variable: Use inverse operations (addition, subtraction, multiplication, or division) to get the variable on one side of the equation.
- Solve for the Variable: Perform the required operations to find the value of the variable.
Example: Solve $ 3x + 5 = 11 $.
Solution:
$ 3x + 5 = 11 $
$ 3x = 11 – 5 $
$ 3x = 6 $
$ x = \frac{6}{3} $
$ x = 2 $
2. What is a Linear Inequality?
A linear inequality is similar to a linear equation but involves an inequality symbol ($<$, $>$, $ \leq $, or $ \geq $) instead of an equal sign. The general form is $ ax + b < c $, where $ x $ is the variable and $ a $, $ b $, and $ c $ are constants.
For example:
- $ 2x + 3 < 7 $ is a linear inequality.
- $ 5x – 1 \geq 9 $ is another linear inequality.
Steps to Solve Linear Inequalities:
- Simplify: Simplify both sides of the inequality.
- Isolate the Variable: Use inverse operations to move the variable to one side.
- Solve for the Variable: Perform the necessary operations.
- Flip the Sign: When multiplying or dividing both sides of an inequality by a negative number, remember to flip the inequality sign.
Example: Solve $ 2x + 3 < 7 $.
Solution:
$ 2x + 3 < 7 $
$ 2x < 7 – 3 $
$ 2x < 4 $
$ x < \frac{4}{2} $
$ x < 2 $
3. Solving Systems of Linear Equations
A system of linear equations consists of two or more linear equations with the same variables. The solution is the set of values that satisfy all the equations simultaneously.
Methods to Solve Systems of Equations:
- Substitution: Solve one equation for one variable and substitute it into the other equation.
- Elimination: Add or subtract equations to eliminate one variable, then solve for the remaining variable.
- Graphing: Graph both equations and find the point where they intersect.
Example: Solve the system:
$ x + y = 5 $
$ 2x – y = 3 $
Solution using substitution:
From the first equation:
$ y = 5 – x $
Substitute this into the second equation:
$ 2x – (5 – x) = 3 $
$ 2x – 5 + x = 3 $
$ 3x – 5 = 3 $
$ 3x = 3 + 5 $
$ 3x = 8 $
$ x = \frac{8}{3} $
Now, substitute $ x = \frac{8}{3} $ into $ y = 5 – x $:
$ y = 5 – \frac{8}{3} $
$ y = \frac{15}{3} – \frac{8}{3} $
$ y = \frac{7}{3} $
Thus, the solution is $ x = \frac{8}{3} $ and $ y = \frac{7}{3} $.
4. Graphing Linear Inequalities
Graphing linear inequalities involves shading a region on a graph that represents all the possible solutions. To graph a linear inequality:
- First, graph the boundary line (the equation obtained by replacing the inequality sign with an equal sign).
- Determine whether to use a solid line (for $ \leq $ or $ \geq $) or a dashed line (for $ < $ or $ > $).
- Test a point that is not on the boundary line to decide which side of the line to shade.
Example: Graph the inequality $ 3x – y \geq 4 $.
Solution:
- Graph the boundary line $ 3x – y = 4 $.
- Since the inequality symbol is $ \geq $, use a solid line.
- Test the point $ (0, 0) $:
$ 3(0) – (0) = 0 $, which is not greater than or equal to 4.
So, shade the region above the line.
5. 100 Examples of Linear Equations and Inequalities
Examples 1 to 10
Example 1: Solve the equation $ 4x + 2 = 10 $.
Solution:
$ 4x + 2 = 10 $
$ 4x = 10 – 2 $
$ 4x = 8 $
$ x = \frac{8}{4} $
$ x = 2 $
Example 2: Solve the inequality $ 3x – 5 < 10 $.
Solution:
$ 3x – 5 < 10 $
$ 3x < 10 + 5 $
$ 3x < 15 $
$ x < \frac{15}{3} $
$ x < 5 $
Examples 91 to 100
Example 91: Solve the system of equations:
$ 3x + 4y = 20 $
$ 2x – y = 7 $
Solution using substitution:
From the second equation:
$ y = 2x – 7 $.
Substitute into the first equation:
$ 3x + 4(2x – 7) = 20 $
$ 3x + 8x – 28 = 20 $
$ 11x – 28 = 20 $
$ 11x = 20 + 28 $
$ 11x = 48 $
$ x = \frac{48}{11} $.
Substitute $ x = \frac{48}{11} $ into $ y = 2x – 7 $:
$ y = 2\left(\frac{48}{11}\right) – 7 $
$ y = \frac{96}{11} – 7 $
$ y = \frac{96}{11} – \frac{77}{11} $
$ y = \frac{19}{11} $.
Thus, $ x = \frac{48}{11} $ and $ y = \frac{19}{11} $.
Example 92: Solve the inequality $ 5x + 9 \geq 24 $.
Solution:
$ 5x + 9 \geq 24 $
$ 5x \geq 24 – 9 $
$ 5x \geq 15 $
$ x \geq \frac{15}{5} $
$ x \geq 3 $
Example 93: Solve the system of equations:
$ 4x + 6y = 22 $
$ 3x – 2y = 5 $
Solution using elimination:
Multiply the second equation by 3 and the first equation by 2:
$ 3(3x – 2y = 5) \rightarrow 9x – 6y = 15 $
$ 2(4x + 6y = 22) \rightarrow 8x + 12y = 44 $.
Add both equations:
$ 9x – 6y + 8x + 12y = 15 + 44 $
$ 17x + 6y = 59 $
$ x = \frac{59}{17} $.
Substitute $ x = \frac{59}{17} $ into $ 4x + 6y = 22 $:
$ 4\left(\frac{59}{17}\right) + 6y = 22 $
$ \frac{236}{17} + 6y = 22 $
$ 6y = 22 – \frac{236}{17} $
$ y = \frac{374}{17} – \frac{236}{17} $
$ y = \frac{138}{17} $.
Thus, $ x = \frac{59}{17} $ and $ y = \frac{138}{17} $.
Example 94: Solve the inequality $ 4x – 6 < 16 $.
Solution:
$ 4x – 6 < 16 $
$ 4x < 16 + 6 $
$ 4x < 22 $
$ x < \frac{22}{4} $
$ x < 5.5 $
Example 95: Solve the system of equations:
$ 3x + 2y = 17 $
$ 5x – y = 14 $
Solution using substitution:
From the second equation:
$ y = 5x – 14 $.
Substitute into the first equation:
$ 3x + 2(5x – 14) = 17 $
$ 3x + 10x – 28 = 17 $
$ 13x – 28 = 17 $
$ 13x = 17 + 28 $
$ 13x = 45 $
$ x = \frac{45}{13} $.
Substitute $ x = \frac{45}{13} $ into $ y = 5x – 14 $:
$ y = 5\left(\frac{45}{13}\right) – 14 $
$ y = \frac{225}{13} – 14 $
$ y = \frac{225}{13} – \frac{182}{13} $
$ y = \frac{43}{13} $.
Thus, $ x = \frac{45}{13} $ and $ y = \frac{43}{13} $.
Example 96: Solve the inequality $ 7x + 5 \geq 30 $.
Solution:
$ 7x + 5 \geq 30 $
$ 7x \geq 30 – 5 $
$ 7x \geq 25 $
$ x \geq \frac{25}{7} $
$ x \geq 3.57 $
Example 97: Solve the system of equations:
$ 4x + 3y = 13 $
$ 2x – y = 4 $
Solution using elimination:
Multiply the second equation by 3:
$ 3(2x – y = 4) \rightarrow 6x – 3y = 12 $.
Now add both equations:
$ (4x + 3y) + (6x – 3y) = 13 + 12 $
$ 10x = 25 $
$ x = \frac{25}{10} = 2.5 $.
Substitute $ x = 2.5 $ into $ 4x + 3y = 13 $:
$ 4(2.5) + 3y = 13 $
$ 10 + 3y = 13 $
$ 3y = 13 – 10 $
$ 3y = 3 $
$ y = 1 $.
Thus, $ x = 2.5 $ and $ y = 1 $.
Example 98: Solve the inequality $ 5x – 8 < 17 $.
Solution:
$ 5x – 8 < 17 $
$ 5x < 17 + 8 $
$ 5x < 25 $
$ x < \frac{25}{5} $
$ x < 5 $
Example 99: Solve the system of equations:
$ 2x + 5y = 22 $
$ 3x – y = 4 $
Solution using substitution:
From the second equation:
$ y = 3x – 4 $.
Substitute into the first equation:
$ 2x + 5(3x – 4) = 22 $
$ 2x + 15x – 20 = 22 $
$ 17x – 20 = 22 $
$ 17x = 22 + 20 $
$ 17x = 42 $
$ x = \frac{42}{17} $.
Substitute $ x = \frac{42}{17} $ into $ y = 3x – 4 $:
$ y = 3\left(\frac{42}{17}\right) – 4 $
$ y = \frac{126}{17} – 4 $
$ y = \frac{126}{17} – \frac{68}{17} $
$ y = \frac{58}{17} $.
Thus, $ x = \frac{42}{17} $ and $ y = \frac{58}{17} $.
Example 100: Solve the inequality $ 3x + 2 > 11 $.
Solution:
$ 3x + 2 > 11 $
$ 3x > 11 – 2 $
$ 3x > 9 $
$ x > \frac{9}{3} $
$ x > 3 $
