Grade 8

Simultaneous equations: practice problems

A pair of equations in two unknowns is solved by eliminating one letter to get a single linear equation.

Basic, Standard, Advanced: Grade 8 · Term 1

BasicStandardAdvancedMix

Elimination

Make the coefficients of one letter equal in size, then add or subtract the equations.

\begin{cases} 3x + 4y = 10 \\ 2x - 3y = 1 \end{cases}

Multiply the first equation by 2 and the second by 3 to get $6x$ in both, then subtract: $17y = 17$ , so $y = 1$ and $x = 2$ .

Tip

Same signs: subtract. Opposite signs: add.

Substitution

If one equation is already $y = \cdots$ or $x = \cdots$ , substitute it into the other.

\begin{cases} y = 2x + 8 \\ x - 4y = -18 \end{cases} \;\Rightarrow\; x - 4(2x + 8) = -18 \;\Rightarrow\; x = -2,\ y = 4

Other forms

Brackets: expand and rewrite as $ax + by = c$ .
Fractions: multiply both sides by the LCM of the denominators.
Decimals: multiply by 10 or 100.
$A = B = C$ : solve the pair $A = C$ and $B = C$ .

Word problems

Choose what $x$ and $y$ stand for, find two equal quantities, solve, and check the answer makes sense. For shopping problems use the total number and the total cost; for speed problems use the total distance and the total time.

Q1BasicGr 8 · Term 1Simultaneous equations

Solve the simultaneous equations.

\begin{cases} 2x + 5y = -13 & \cdots\ (1) \\ -2x - 4y = 12 & \cdots\ (2) \end{cases}

Answer

x = -4,\ y = -1

Solution

Eliminate x: (1) + (2).
$\begin{array}{lrcrcr} & 2x & + & 5y & = & -13 \\ +)\ & -2x & - & 4y & = & 12 \\ \hline & & & y & = & -1 \end{array}$
Substitute $y = -1$ into (1).
$2x + 5 \times (-1) = -13 \quad\Rightarrow\quad 2x = -8 \quad\Rightarrow\quad x = -4$

Q2StandardGr 8 · Term 1Simultaneous equations

Solve the simultaneous equations.

x - 2y = x - 5y - 3 = -2

Answer

x = -4,\ y = -1

Solution

Write it as $A = C$ and $B = C$ and simplify.
$\begin{cases} x - 2y = -2 & \cdots\ (1) \\ x - 5y = 1 & \cdots\ (2) \end{cases}$
Eliminate x: (1) − (2).
$\begin{array}{lrcrcr} & x & - & 2y & = & -2 \\ -)\ & x & - & 5y & = & 1 \\ \hline & & & 3y & = & -3 \end{array}$
Divide both sides by $3$ .
$y = -1$
Substitute $y = -1$ into (1).
$x - 2 \times (-1) = -2 \quad\Rightarrow\quad x = -4$

Q3AdvancedGr 8 · Term 1Simultaneous equations

The digits of a two-digit number add up to 11. If the digits are reversed, the new number is 63 greater than the original. Find the original number.

Answer

29

Solution

Let the tens digit be $x$ and the units digit $y$ .
$\begin{cases} x + y = 11 & \cdots\ (1) \\ 10y + x = 10x + y + 63 & \cdots\ (2) \end{cases}$
Simplify (2):
$-9x + 9y = 63 \quad\Rightarrow\quad -x + y = 7$
Solve with (1):
$x = 2,\ y = 9$
So the number is
$29$

Q1BasicGr 8 · Term 1Simultaneous equations

Solve the simultaneous equations.

\begin{cases} x = 4y + 11 & \cdots\ (1) \\ -2x - 6y = 34 & \cdots\ (2) \end{cases}

Q2StandardGr 8 · Term 1Simultaneous equations

Solve the simultaneous equations.

\begin{cases} 2(x + y) - 5y = 0 & \cdots\ (1) \\ -3x - 5(y - 3) = -4 & \cdots\ (2) \end{cases}

Q3AdvancedGr 8 · Term 1Simultaneous equations

There are 22 coins, all 10-yen and 50-yen coins, worth 660 yen in total. How many of each coin are there?

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