Equations to Tidy First and Choosing a Method

Put simultaneous equations with terms on both sides, brackets, fractions or decimals into the form $ax + by = c$ ; split the form $A = B = C$ ; choose between substitution and elimination; and use a solution once you have it.

Before you start

Substitution (lesson 2.6) and elimination (lesson 2.7) both start from two equations in the form $ax + by = c$ .
To clear fractions, multiply every term by the lowest common multiple of the denominators: $\frac{x}{2} + \frac{y}{3} = 4$ times $6$ is $3x + 2y = 24$ (Secondary 1).
Expanding: $-3(x - y) = -3x + 3y\text{.}$

Tidying Each Equation First

Tidy each equation into the form ax + by = c

Terms on both sides: move the $x$ and $y$ terms to the left and the numbers to the right.

Brackets: expand them, then collect like terms.

Fractions: multiply every term by the lowest common multiple of the denominators.

Decimals: multiply every term by $10$ (or $100$ ) so that every coefficient is a whole number.

Then solve by substitution or elimination, and check in the equations as they were given.

Example 1: Terms on both sides

Solve $2x = 3y + 1$ and $5x - 6 = 4y$ .

Solution

First equation: subtract $3y$ from both sides: $2x - 3y = 1$ $\quad$ (1)

Second equation: subtract $4y$ and add $6$ : $5x - 4y = 6$ $\quad$ (2)

Eliminate $y$ (lowest common multiple $12$ ). (1) $\times 4$ : $8x - 12y = 4$ . (2) $\times 3$ : $15x - 12y = 18$ .

Subtract: $7x = 14$ , so $x = 2$ . Substitute into (1): $4 - 3y = 1$ , so $y = 1$ .

Check in the given equations: $2(2) = 4$ and $3(1) + 1 = 4$ , true; $5(2) - 6 = 4$ and $4(1) = 4$ , true.

$x = 2$ , $y = 1$

Example 2: Brackets

Solve $2(x + y) - 3(x - y) = 9$ and $4(x - 1) - (y + 2) = -4\text{.}$

Solution

First: $2x + 2y - 3x + 3y = 9\text{,}$ so $-x + 5y = 9$ $\quad$ (1)

Second: $4x - 4 - y - 2 = -4$ , so $4x - y - 6 = -4$ and $4x - y = 2$ $\quad$ (2)

The minus sign in front of $(y + 2)$ changes both signs: $-(y + 2) = -y - 2$ .

Eliminate $x$ : (1) $\times 4$ : $-4x + 20y = 36$ . Add (2): $19y = 38$ , so $y = 2$ .

Substitute into (2): $4x - 2 = 2$ , so $x = 1$ .

Check in the given equations: $2(1 + 2) - 3(1 - 2) = 6 + 3 = 9\text{,}$ true; $4(1 - 1) - (2 + 2) = 0 - 4 = -4\text{,}$ true.

$x = 1$ , $y = 2$

Example 3: Fractions

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

Solution

First: the lowest common multiple of $2$ and $3$ is $6$ . Multiply every term by $6$ : $3x + 2y = 24$ $\quad$ (1)

Second: multiply every term by $3$ : $x - 3y = -3$ $\quad$ (2)

In (2), $x$ has coefficient $1$ : $x = 3y - 3$ . Substitute into (1): $3(3y - 3) + 2y = 24$ , so $11y - 9 = 24$ , $11y = 33$ , $y = 3$ .

Then $x = 3(3) - 3 = 6$ .

Check in the given equations: $\frac{6}{2} + \frac{3}{3} = 3 + 1 = 4\text{,}$ true; $\frac{6}{3} - 3 = 2 - 3 = -1\text{,}$ true.

$x = 6$ , $y = 3$

Example 4: Decimals

Solve $0.3x + 0.2y = 1.3$ and $0.5x - 0.1y = 2.6$ .

Solution

Multiply every term of each equation by $10$ : $3x + 2y = 13$ $\quad$ (1) and $5x - y = 26$ $\quad$ (2)

From (2): $y = 5x - 26$ . Substitute into (1): $3x + 2(5x - 26) = 13\text{,}$ so $13x - 52 = 13$ , $13x = 65$ , $x = 5$ .

Then $y = 25 - 26 = -1$ .

Check in the given equations: $0.3(5) + 0.2(-1) = 1.5 - 0.2 = 1.3\text{,}$ true; $0.5(5) - 0.1(-1) = 2.5 + 0.1 = 2.6\text{,}$ true.

$x = 5$ , $y = -1$

The Form A = B = C

Sometimes two equations are written as one line with two equals signs, such as $2x + y = x - y + 5 = 7\text{.}$ It says that all three expressions are equal. Split it into two equations, each with one equals sign; any two of the three pairs will do, and the simplest choice usually uses the number on its own.

Example 5: Splitting into two equations

Solve $2x + y = x - y + 5 = 7\text{.}$

Solution

Split: $2x + y = 7$ $\quad$ (1) and $x - y + 5 = 7$ , so $x - y = 2$ $\quad$ (2)

Add (1) and (2): $3x = 9$ , so $x = 3$ . Then from (2), $3 - y = 2$ , so $y = 1$ .

Check all three expressions: $2(3) + 1 = 7$ ; $3 - 1 + 5 = 7$ ; and $7$ . All equal.

$x = 3$ , $y = 1$

Choosing a Method

Both methods give the same answer. Choose the one that will be quicker and safer for the pair in front of you.

If the pair looks like this	Use	Why
One equation already gives $x$ or $y$ , such as $y = 4 - x$	substitution	nothing to rearrange
A variable has coefficient $1$ or $-1$ , such as $2x - y = 4$	substitution, or elimination	the subject is easy to make; elimination is also quick
A variable has the same or opposite coefficients, such as $3x + 4y = 17$ and $3x - 2y = 5$	elimination	one subtraction or addition removes it
No coefficient is $1$ or $-1$ , such as $4x + 3y = 18$ and $5x + 2y = 19$	elimination	substitution would bring fractions

Example 6: The same pair both ways

Solve $2x + 3y = 12$ and $2x - y = 4$ (a) by elimination, (b) by substitution.

Solution

$2x + 3y = 12$ $\quad$ (1), $\quad$ $2x - y = 4$ $\quad$ (2)

(a) Both have $2x$ . Subtract (2) from (1): $3y - (-y) = 12 - 4$ , so $4y = 8$ and $y = 2$ . Then $2x - 2 = 4$ , so $x = 3$ .

(b) From (2): $y = 2x - 4$ . Substitute into (1): $2x + 3(2x - 4) = 12$ , so $8x - 12 = 12$ , $8x = 24$ , $x = 3$ . Then $y = 6 - 4 = 2$ .

Check: $6 + 6 = 12$ ; $6 - 2 = 4$ . Both methods give the same answer; here elimination took one step fewer.

$x = 3$ , $y = 2$ by either method

Unknown Constants, and Using a Solution

If you are told the solution and the equations contain unknown constants, substitute the solution. The constants then satisfy a new pair of simultaneous equations.

Example 7: Finding a and b

$x = 2$ , $y = 1$ is the solution of the simultaneous equations $ax + by = 7$ and $ax - by = 1$ . Find $a$ and $b$ .

Solution

Substitute $x = 2$ and $y = 1$ : $2a + b = 7$ $\quad$ (1) and $2a - b = 1$ $\quad$ (2)

Add: $4a = 8$ , so $a = 2$ . Then $4 + b = 7$ , so $b = 3$ .

Check: the equations are $2x + 3y = 7$ and $2x - 3y = 1$ . At $(2, 1)$ : $4 + 3 = 7$ and $4 - 3 = 1$ , both true.

$a = 2$ , $b = 3$

Example 8: Using the solution

Solve $3x + y = 11$ and $x - 2y = -8$ . Hence find the value of (a) $x^{2} + y$ , (b) $3x - y$ .

Solution

From the first: $y = 11 - 3x$ . Substitute: $x - 2(11 - 3x) = -8\text{,}$ so $x - 22 + 6x = -8$ , $7x = 14$ , $x = 2$ . Then $y = 11 - 6 = 5$ .

Check: $6 + 5 = 11$ ; $2 - 10 = -8$ .

(a) $x^{2} + y = 2^{2} + 5 = 4 + 5 = 9\text{.}$

(b) $3x - y = 6 - 5 = 1$ .

$x = 2$ , $y = 5$ ; (a) $9$ (b) $1$

Common Mistakes

Multiplying only the fractions. $\frac{x}{2} + \frac{y}{3} = 4$ times $6$ is $3x + 2y = 24$ , not $3x + 2y = 4$ : the $4$ is multiplied too.
A minus sign in front of a bracket. $-(y + 2) = -y - 2$ , and $-3(x - y) = -3x + 3y\text{.}$
Multiplying decimals by $10$ on one side only, or by $10$ when $100$ is needed ( $0.05x$ needs $100$ ).
Splitting $A = B = C$ into one equation only, or into $A = B$ and $A = B$ again. You need two different equations.
Checking in the tidied equations only. A mistake made while tidying is caught only by the equations as they were given.

Practice Questions

Work each question on paper before you open its solution, and check in the equations as given. The questions get harder as you go.

Question 1

Solve $3x = 2y + 4$ and $x + 1 = y$ .

ShowHide solution

The second already gives $y = x + 1$ . Substitute into the first: $3x = 2(x + 1) + 4$ , so $3x = 2x + 6$ and $x = 6$ . Then $y = 7$ .

Check: $3(6) = 18$ and $2(7) + 4 = 18$ ; $6 + 1 = 7$ .

$x = 6$ , $y = 7$

Question 2

Solve $3(x + 2y) - 2(x - y) = 13$ and $2(x - 1) = y + 7$ .

ShowHide solution

First: $3x + 6y - 2x + 2y = 13\text{,}$ so $x + 8y = 13$ . Second: $2x - 2 = y + 7$ , so $2x - y = 9$ .

From the second: $y = 2x - 9$ . Substitute: $x + 8(2x - 9) = 13$ , so $17x - 72 = 13$ , $17x = 85$ , $x = 5$ . Then $y = 1$ .

Check: $3(7) - 2(4) = 21 - 8 = 13\text{;}$ $2(4) = 8$ and $1 + 7 = 8$ .

$x = 5$ , $y = 1$

Question 3

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

ShowHide solution

First $\times 6$ : $2x + 3y = 24$ . Second $\times 4$ : $2x - y = 8$ .

Subtract: $4y = 16$ , so $y = 4$ . Then $2x - 4 = 8$ , so $x = 6$ .

Check: $2 + 2 = 4$ ; $3 - 1 = 2$ .

$x = 6$ , $y = 4$

Question 4

Solve $\frac{x + 1}{2} + \frac{y - 1}{3} = 3$ and $\frac{x + y}{7} = 1$ .

ShowHide solution

First $\times 6$ : $3(x + 1) + 2(y - 1) = 18\text{,}$ so $3x + 3 + 2y - 2 = 18$ and $3x + 2y = 17$ .

Second $\times 7$ : $x + y = 7$ . Multiply by $2$ : $2x + 2y = 14$ . Subtract from $3x + 2y = 17$ : $x = 3$ . Then $y = 4$ .

Check: $\frac{4}{2} + \frac{3}{3} = 2 + 1 = 3\text{;}$ $\frac{7}{7} = 1$ .

$x = 3$ , $y = 4$

Question 5

Solve $0.2x + 0.5y = 1.6$ and $0.3x - 0.1y = -1$ .

ShowHide solution

Multiply each by $10$ : $2x + 5y = 16$ and $3x - y = -10$ .

From the second: $y = 3x + 10$ . Substitute: $2x + 5(3x + 10) = 16\text{,}$ so $17x + 50 = 16$ , $17x = -34$ , $x = -2$ . Then $y = -6 + 10 = 4$ .

Check: $0.2(-2) + 0.5(4) = -0.4 + 2 = 1.6\text{;}$ $0.3(-2) - 0.1(4) = -0.6 - 0.4 = -1\text{.}$

$x = -2$ , $y = 4$

Question 6

Solve $x + 2y = 3x - y + 1 = 10\text{.}$

ShowHide solution

Split: $x + 2y = 10$ and $3x - y + 1 = 10$ , so $3x - y = 9$ .

From the second: $y = 3x - 9$ . Substitute: $x + 2(3x - 9) = 10$ , so $7x - 18 = 10$ , $7x = 28$ , $x = 4$ . Then $y = 3$ .

Check: $4 + 6 = 10$ ; $12 - 3 + 1 = 10$ .

$x = 4$ , $y = 3$

Question 7

For each pair, choose a method, say why, and solve. (a) $x = 2y + 1$ and $3x + 4y = 15$ (b) $5x + 2y = 11$ and $5x - 3y = 1$ (c) $3x + 2y = 12$ and $4x + 5y = 23$

ShowHide solution

(a) Substitution: $x$ is already the subject. $3(2y + 1) + 4y = 15$ , so $10y + 3 = 15$ , $y = 1.2$ . Then $x = 2(1.2) + 1 = 3.4\text{.}$ Check: $2(1.2) + 1 = 3.4$ ; $3(3.4) + 4(1.2) = 10.2 + 4.8 = 15\text{.}$

(b) Elimination: both have $5x$ . Subtract: $2y - (-3y) = 11 - 1\text{,}$ so $5y = 10$ , $y = 2$ . Then $5x + 4 = 11$ , $x = 1.4$ . Check: $7 + 4 = 11$ ; $7 - 6 = 1$ .

(c) Elimination: no coefficient is $1$ . Eliminate $y$ : first $\times 5$ : $15x + 10y = 60$ ; second $\times 2$ : $8x + 10y = 46$ . Subtract: $7x = 14$ , $x = 2$ . Then $6 + 2y = 12$ , $y = 3$ . Check: $6 + 6 = 12$ ; $8 + 15 = 23$ .

(a) $x = 3.4$ , $y = 1.2$ (b) $x = 1.4$ , $y = 2$ (c) $x = 2$ , $y = 3$

Question 8

Solve $x + 3y = 7$ and $2x - y = 7$ by substitution and again by elimination, and show that the answers agree.

ShowHide solution

Substitution: $x = 7 - 3y$ . Then $2(7 - 3y) - y = 7$ , so $14 - 7y = 7$ , $y = 1$ , and $x = 4$ .

Elimination: first $\times 2$ : $2x + 6y = 14$ . Subtract $2x - y = 7$ : $7y = 7$ , $y = 1$ ; then $x = 7 - 3 = 4$ .

Check: $4 + 3 = 7$ ; $8 - 1 = 7$ .

$x = 4$ , $y = 1$ both ways

Question 9

The simultaneous equations $ax + by = 11$ and $bx + ay = -1$ have the solution $x = 3$ , $y = -1$ . Find $a$ and $b$ .

ShowHide solution

Substitute: $3a - b = 11$ and $3b - a = -1$ , that is $-a + 3b = -1$ .

From the first: $b = 3a - 11$ . Substitute: $-a + 3(3a - 11) = -1\text{,}$ so $8a - 33 = -1$ , $8a = 32$ , $a = 4$ . Then $b = 1$ .

Check: $4(3) + 1(-1) = 11$ ; $1(3) + 4(-1) = -1\text{.}$

$a = 4$ , $b = 1$

Question 10

Solve $2x + y = 7$ and $x - y = 5$ . Hence find the value of (a) $x^{2} + y^{2}$ , (b) $2x - 3y$ .

ShowHide solution

Add: $3x = 12$ , so $x = 4$ . Then $y = 7 - 8 = -1$ . Check: $8 - 1 = 7$ ; $4 + 1 = 5$ .

(a) $4^{2} + (-1)^{2} = 16 + 1 = 17\text{.}$ (b) $2(4) - 3(-1) = 8 + 3 = 11\text{.}$

$x = 4$ , $y = -1$ ; (a) $17$ (b) $11$

Question 11

Twice a number added to a third of a second number is $17$ . A quarter of the first number added to the second number is $5$ . Find the two numbers.

ShowHide solution

Let the numbers be $x$ and $y$ : $2x + \frac{y}{3} = 17$ and $\frac{x}{4} + y = 5$ .

Clear the fractions: $6x + y = 51$ and $x + 4y = 20$ .

From the first: $y = 51 - 6x$ . Substitute: $x + 4(51 - 6x) = 20$ , so $x + 204 - 24x = 20$ , $-23x = -184$ , $x = 8$ . Then $y = 51 - 48 = 3$ .

Check: $16 + 1 = 17$ ; $2 + 3 = 5$ .

The numbers are $8$ and $3$ .

Question 12

At a fruit stall, mangoes cost $ $1.20$ each and apples cost $ $0.50$ each. Siti buys 10 fruits and pays $ $7.80$ . How many of each does she buy?

ShowHide solution

Let her buy $x$ mangoes and $y$ apples: $x + y = 10$ and $1.2x + 0.5y = 7.8$ .

Multiply the second by $10$ : $12x + 5y = 78$ . From the first, $y = 10 - x$ : $12x + 50 - 5x = 78$ , so $7x = 28$ , $x = 4$ . Then $y = 6$ .

Check: $4 + 6 = 10$ ; $4(1.20) + 6(0.50) = 4.80 + 3.00 = 7.80\text{.}$

She buys 4 mangoes and 6 apples.

Key Points

Tidy each equation into the form $ax + by = c$ before solving: collect terms, expand brackets, clear fractions and decimals.
Clear fractions by multiplying every term by the lowest common multiple of the denominators; clear decimals by $10$ or $100$ .
$A = B = C$ gives two equations, for example $A = C$ and $B = C$ .
Substitution suits a pair where a variable is already, or easily, the subject; elimination suits matching or larger coefficients. Both give the same answer.
Given the solution, substitute it to find unknown constants: they satisfy a new pair of equations.
Check in the equations as they were given, not only in the tidied ones.