Year 10 and 10A Algebra, equations and linear relationships: free practice test with answers

Sample algebra, equations and linear relationships questions with answers

Try each one before you open the answer.

1. Review of algebra

   Simplify $\frac{3x}{8} + \frac{x}{8}.$

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   Key idea: When denominators are the same, add the numerators and keep the denominator.

   1. Add the numerators: $3x + x = 4x$
   2. Write over the common denominator: $\frac{4x}{8}$
   3. Simplify by dividing numerator and denominator by 4: $\frac{x}{2}$

   Answer: $\dfrac{x}{2}$

2. Solving linear equations

   Solve $\dfrac{x + 3}{4} = 7$.

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   Key idea: Multiply both sides by the denominator to eliminate the fraction, then solve.

   1. Multiply both sides by $4$:\

   $$\begin{gathered}4 \times \frac{x+3}{4} = 7 \times 4 \\ x + 3 = 28\end{gathered}$$

   2. Subtract $3$ from both sides:\

   $$x = 25$$

   3. Check: $\dfrac{25 + 3}{4} = \dfrac{28}{4} = 7$ ✓

   Answer: $x = 25$

3. Linear equations involving more complex algebraic fractions

   Solve $\frac{x + 2}{3} - \frac{x - 1}{4} = 2$.

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   Key idea: Multiply every term by the lowest common denominator, 12, to clear the fractions.

   1. $4(x + 2) - 3(x - 1) = 24$
   2. $4x + 8 - 3x + 3 = 24$
   3. $x + 11 = 24$, so $x = 13$

   Answer: $x = 13$

4. Graphing straight lines

   Consider the line $3x - 4y = 24$.

   (a) Find the $x$-intercept and the $y$-intercept.

   (b) Find the gradient of the line.

   (c) Describe how to sketch the line using your answers.

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   Key idea: To find an intercept, set the other variable to zero. Rearranging to $y = mx + c$ gives the gradient.

   (a)

   1. $x$-intercept: let $y = 0$, so $3x = 24$ and $x = 8$. The point is $(8, 0)$.
   2. $y$-intercept: let $x = 0$, so $-4y = 24$ and $y = -6$. The point is $(0, -6)$.

   (b)

   1. Rearrange: $-4y = -3x + 24$, so $y = \frac{3}{4}x - 6$.
   2. The gradient is $\frac{3}{4}$.
   3. Check with the intercepts: $\frac{\text{rise}}{\text{run}} = \frac{0 - (-6)}{8 - 0} = \frac{6}{8} = \frac{3}{4}$ ✓

   (c)

   1. Draw axes, plot $(8, 0)$ and $(0, -6)$, then rule a straight line through both points and label it $3x - 4y = 24$.
   2. The line rises from left to right, which matches the positive gradient.

   Answer: (a) $x$-intercept $(8, 0)$, $y$-intercept $(0, -6)$ (b) $\frac{3}{4}$ (c) Plot the two intercepts and rule a line through them.

5. Length and midpoint of a line segment

   For the points $A(-3, 4)$ and $B(5, -2)$, find:

   (a) the length of the line segment $AB$

   (b) the midpoint of $AB$.

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   Key idea: Length $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ and midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

   (a)

   1. $x_2 - x_1 = 5 - (-3) = 8$ and $y_2 - y_1 = -2 - 4 = -6$.
   2. $AB = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

   (b)

   1. $M = \left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right) = (1, 1)$

   Answer: (a) 10 units (b) $(1, 1)$

6. Parallel and perpendicular lines

   Line $\ell$ has equation $y = 3x - 5$. Write down the gradient of any line that is (a) parallel to $\ell$, and (b) perpendicular to $\ell$.

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   Key idea: Parallel lines have equal gradients. Perpendicular lines have gradients whose product is $-1$, so the perpendicular gradient is the negative reciprocal.

   1. The gradient of $\ell$ is $m = 3$.
   2. (a) Parallel gradient: A parallel line has the same gradient, so $m_{\parallel} = 3$.
   3. (b) Perpendicular gradient: Flip and negate: $m_{\perp} = -\dfrac{1}{3}$.
   4. Check: $3 \times \left(-\dfrac{1}{3}\right) = -1$ ✓

   Answer: Parallel gradient $= 3$; perpendicular gradient $= -\dfrac{1}{3}$.

7. Simultaneous equations using elimination

   Solve the simultaneous equations:\

   $$\begin{gathered}3x + 2y = 16 \\ 5x - 2y = 8\end{gathered}$$
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   Key idea: Add the equations to eliminate $y$, since the $y$-coefficients are opposites.

   1. Add the two equations:\

   $$\begin{aligned}(3x + 2y) + (5x - 2y) &= 16 + 8 \\ 8x &= 24 \\ x &= 3\end{aligned}$$

   2. Substitute $x = 3$ into $3x + 2y = 16$:\

   $$\begin{aligned}9 + 2y &= 16 \\ 2y &= 7 \\ y &= 3.5\end{aligned}$$

   3. Check in $5x - 2y = 8$: $5(3) - 2(3.5) = 15 - 7 = 8$ ✓

   Answer: $x = 3,\; y = 3.5$

8. Further applications of simultaneous equations

   Two friends, Aiden and Brianna, buy items at a school canteen. Aiden buys 2 pies and 1 drink and pays $8.70. Brianna buys 1 pie and 1 drink and pays $5.20. Find the cost of one pie and one drink.

   Show the answer

   Key idea: Define variables, write two equations from the given information, then solve.

   1. Let $p$ = cost of one pie (dollars) and $d$ = cost of one drink (dollars). Write the equations:\

   $$\begin{aligned}2p + d &= 8.70 (1)\\ p + d &= 5.20 (2)\end{aligned}$$

   2. Subtract equation (2) from equation (1) to eliminate $d$:\

   $$\begin{aligned}(2p + d) - (p + d) &= 8.70 - 5.20 \\ p &= 3.50\end{aligned}$$

   3. Substitute $p = 3.50$ into equation (2):\

   $$\begin{aligned}3.50 + d &= 5.20 \\ d &= 1.70\end{aligned}$$

   4. Check in equation (1): $2(3.50) + 1.70 = 7.00 + 1.70 = 8.70$ ✓

   Answer: One pie costs $3.50 and one drink costs $1.70.

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 10 algebraic fractions worksheet
• Year 10 substituting and rearranging formulas worksheet
• Year 10 linear equations worksheet
• Year 10 linear equations with algebraic fractions worksheet
• Year 10 linear inequalities worksheet
• Year 10 coordinate geometry worksheet
• Year 10 simultaneous equations worksheet
• Year 10A linear and non-linear simultaneous equations worksheet