Quiz: Linear Equation Practice Problems with One or Two Variables (set 1)

This quiz focuses on linear equations with one and two variables and is primarily designed for high school students. However, anyone interested in practicing these concepts is welcome to take the quiz! To learn more about solving linear equations with one or two variables, feel free to watch the video tutorial provided in this post. Try to solve the questions without external help. If you don’t get all the answers correct on your first attempt, don’t worry—you can retake the quiz to aim for a perfect score. If you notice any errors in the questions or answers, please leave a comment so we can address them. Enjoy the quiz and good luck!

1. 
Solve for $x$: $\displaystyle \frac{2x - 3}{4} + \frac{3x + 1}{2} = \frac{5x - 1}{6}$

2. 
Solve for $x$: $\displaystyle \frac{1}{x-2} + \frac{2}{x+3} = \frac{3}{x^2 + x - 6}$

3. 
Solve for $x$: $\displaystyle |2x - 5| = 7$

4. 
Solve for $x: \\ 3(2x−4)−2(3x+1) = 4(x−5)$

5. 
Solve for $x$: $\displaystyle \frac{3x+2}{5} - \frac{2x−1}{3} = \frac{x+4}{15}$

6. 
Solve for $x$ and $y$: $\begin{cases} 3x + 4y = 10 \\ 5x - 2y = 8 \end{cases} $

7. 
Solve for $x$ and $y$: $\large \begin{cases} \frac{x}{2} + \frac{y}{3} = 4 \\ \frac{x}{3} - \frac{y}{2} = 1 \end{cases}$

8. 
Solve for $x$ and $y$: $\begin{cases} 2x+5y=12 \\ 4x+10y = 24 \end{cases}$

9. 
A boat travels 60 km downstream in 3 hours and takes 5 hours to travel the same distance upstream. Find the speed of the boat in still water and the speed of the current.

10. 
A rectangle has a perimeter of 40 cm. If the length is reduced by 2 cm and the width is increased by 2 cm, the area of the rectangle increases by 4 cm$^2$. Find the original dimensions of the rectangle.

11. 
A shopkeeper sells two types of pens. Type A costs 5 each and Type B costs 8 each. If the shopkeeper sells 50 pens for a total of $340, how many pens of each type were sold?

12. 
The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 less than the original number. Find the original number.

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