Sec 2 · G3 (Express)

Linear equations, inequalities and simultaneous equations

Practise linear equations, inequalities and simultaneous equations with step-by-step help. Free account.

1. Equations with fractions

Multiply every term by the denominator

\displaystyle \dfrac{2x}{3} - 1 = 5

e.g.

\displaystyle \dfrac{2x}{3} - 1 = 5

\displaystyle 2x - 3 = 15

\displaystyle x = 9

Two fractions: multiply by the LCM

\displaystyle \dfrac{x}{2} + \dfrac{x}{5} = 7

e.g.

\displaystyle \dfrac{x}{2} + \dfrac{x}{5} = 7

\displaystyle 5x + 2x = 70

\displaystyle x = 10

📖 Worked example

Solve $\dfrac{x}{6} - \dfrac{x}{3} = -1$ .

Step 1 of 4

Find the LCM of the denominators

\text{LCM of } 6 \text{ and } 3 =

6

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2. Simple inequalities

\displaystyle x > 2

\displaystyle x \leq -1

Solve like an equation

\displaystyle 3x - 4 < 11

e.g.

\displaystyle 3x - 4 < 11

\displaystyle 3x < 15

\displaystyle x < 5

Multiply or divide by a negative number: reverse the inequality sign

\displaystyle -2x \geq 8

e.g.

\displaystyle -2x \geq 8

\displaystyle x \leq -4

📖 Worked example

Solve $-7 - 2x < -9$ .

Step 1 of 2

Collect the x terms on one side and the numbers on the other

-2x < -2

📖 Worked example

Write down the inequality shown on the number line.

Step 1 of 2

Read the circles

A filled circle means the value is included (

\leq

\geq

); an open circle means it is not (

<

>

Practise this →

3. Graphs of linear equations

Equation	Gradient	$y$ -intercept
$y = mx + c$	$m$	$c$
$2x + 3y = 12$ , so $y = -\dfrac{2}{3}x + 4$	$-\dfrac{2}{3}$	$4$
$x = 3$	undefined (vertical line)	none
$y = -2$	$0$ (horizontal line)	$-2$

Where 2x + 3y = 12 cuts the axes

\displaystyle 2x + 3y = 12

$x = 0$ : $3y = 12$ , so $(0, 4)$
$y = 0$ : $2x = 12$ , so $(6, 0)$

📖 Worked example

Find the $y$ -intercept of the line $-2x - y = -8$ .

Step 1 of 3

Make y the subject: move the x term across (it changes sign)

-y = 2x - 8

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4. Simultaneous equations

The lines y = 2x − 1 and x + y = 5 cross at (2, 3), the solution of both equations.

Substitution: y = 2x − 1 into x + y = 5

\displaystyle x + (2x - 1) = 5

e.g.

\displaystyle x + (2x - 1) = 5

\displaystyle 3x - 1 = 5

\displaystyle x = 2

$y = 2(2) - 1$ , which is $3$

Elimination: add to remove y

\displaystyle 3x + 2y = 16 \text{ and } 5x - 2y = 16

e.g.

\displaystyle 8x = 32

\displaystyle x = 4

$3(4) + 2y = 16$ , so $y = 2$

📖 Worked example

Solve the simultaneous equations $-6x - 3y = 24$ and $x + 3y = -14$ .

Step 1 of 4

Number the equations

-6x - 3y = 24

... (1)

x + 3y = -14

... (2)

📖 Worked example

Solve the simultaneous equations $-3x - 3y = -33$ and $y = 3x - 9$ by substitution.

Step 1 of 4

Number the equations

-3x - 3y = -33

... (1)

y = 3x - 9

... (2)

📖 Worked example

3 curry puffs and 1 packet of soya bean milk cost $4.70. 1 curry puff and 2 packets of soya bean milk cost $3.40. Find the cost of one curry puff and the cost of one packet of soya bean milk.

Step 1 of 6

Let one curry puff cost x dollars and one packet of soya bean milk cost y dollars. The first purchase

3x + y = 4.7

... (1)

Where the two equations come from	Equation
two lines meet at $P$	$P$ fits both lines' equations
opposite sides of a rectangle	$AB = DC$ , $AD = BC$
opposite angles of a parallelogram; co-interior angles	$\angle A = \angle C$ ; $\angle A + \angle B = 180^\circ$
a fraction $\dfrac{x}{y}$ that becomes $\dfrac{a}{b}$	cross-multiply: $b(\ldots) = a(\ldots)$

📖 Worked example

The lines $5x + 6y = 62$ and $-2x + y = -1$ meet at the point $P$ . Find the coordinates of $P$ .

Step 1 of 5

P is on both lines, so its x and y fit both equations: solve them together. Number the equations

5x + 6y = 62

... (1)

-2x + y = -1

... (2)

📖 Worked example

In the rectangle $ABCD$ , $AB = (4x + 2y + 5)$ cm, $DC = (2x + 3y + 5)$ cm, $AD = (x + y + 2)$ cm and $BC = 11$ cm. Form two equations in $x$ and $y$ and solve them. Hence find the perimeter of the rectangle.

Step 1 of 6

Opposite sides of a rectangle are equal: AB = DC. Simplify

4x + 2y + 5 = 2x + 3y + 5

2x - y = 0

... (1)

📖 Worked example

When $3$ is added to both the numerator and the denominator of a fraction, the fraction becomes $\dfrac{7}{12}$ . When $3$ is subtracted from both the numerator and the denominator of the original fraction, it becomes $\dfrac{1}{6}$ . Let the fraction be $\dfrac{x}{y}$ . Find the fraction.

Step 1 of 6

Write each fact as an equation, cross-multiply and simplify

\dfrac{x + 3}{y + 3} = \dfrac{7}{12}

12(x + 3) = 7(y + 3)

12x - 7y = -15

... (1)

\dfrac{x - 3}{y - 3} = \dfrac{1}{6}

6(x - 3) = 1(y - 3)

6x - y = 15

... (2)

Practise this →

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Sec 2 · G3 (Express)

Linear equations, inequalities and simultaneous equations

Practise linear equations, inequalities and simultaneous equations with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Equations with fractions

Multiply every term by the denominator

\displaystyle \dfrac{2x}{3} - 1 = 5

e.g.

\displaystyle \dfrac{2x}{3} - 1 = 5

\displaystyle 2x - 3 = 15

\displaystyle x = 9

Two fractions: multiply by the LCM

\displaystyle \dfrac{x}{2} + \dfrac{x}{5} = 7

e.g.

\displaystyle \dfrac{x}{2} + \dfrac{x}{5} = 7

\displaystyle 5x + 2x = 70

\displaystyle x = 10

📖 Worked example

Solve $\dfrac{x}{6} - \dfrac{x}{3} = -1$ .

Step 1 of 4

Find the LCM of the denominators

\text{LCM of } 6 \text{ and } 3 =

6

Practise this →

2. Simple inequalities

\displaystyle x > 2

\displaystyle x \leq -1

Solve like an equation

\displaystyle 3x - 4 < 11

e.g.

\displaystyle 3x - 4 < 11

\displaystyle 3x < 15

\displaystyle x < 5

Multiply or divide by a negative number: reverse the inequality sign

\displaystyle -2x \geq 8

e.g.

\displaystyle -2x \geq 8

\displaystyle x \leq -4

📖 Worked example

Solve $-7 - 2x < -9$ .

Step 1 of 2

Collect the x terms on one side and the numbers on the other

-2x < -2

📖 Worked example

Write down the inequality shown on the number line.

Step 1 of 2

Read the circles

A filled circle means the value is included (

\leq

\geq

); an open circle means it is not (

<

>

Practise this →

3. Graphs of linear equations

Equation	Gradient	$y$ -intercept
$y = mx + c$	$m$	$c$
$2x + 3y = 12$ , so $y = -\dfrac{2}{3}x + 4$	$-\dfrac{2}{3}$	$4$
$x = 3$	undefined (vertical line)	none
$y = -2$	$0$ (horizontal line)	$-2$

Where 2x + 3y = 12 cuts the axes

\displaystyle 2x + 3y = 12

$x = 0$ : $3y = 12$ , so $(0, 4)$
$y = 0$ : $2x = 12$ , so $(6, 0)$

📖 Worked example

Find the $y$ -intercept of the line $-2x - y = -8$ .

Step 1 of 3

Make y the subject: move the x term across (it changes sign)

-y = 2x - 8

Practise this →

4. Simultaneous equations

Substitution: y = 2x − 1 into x + y = 5

\displaystyle x + (2x - 1) = 5

e.g.

\displaystyle x + (2x - 1) = 5

\displaystyle 3x - 1 = 5

\displaystyle x = 2

$y = 2(2) - 1$ , which is $3$

Elimination: add to remove y

\displaystyle 3x + 2y = 16 \text{ and } 5x - 2y = 16

e.g.

\displaystyle 8x = 32

\displaystyle x = 4

$3(4) + 2y = 16$ , so $y = 2$

📖 Worked example

Solve the simultaneous equations $-6x - 3y = 24$ and $x + 3y = -14$ .

Step 1 of 4

Number the equations

-6x - 3y = 24

... (1)

x + 3y = -14

... (2)

📖 Worked example

Solve the simultaneous equations $-3x - 3y = -33$ and $y = 3x - 9$ by substitution.

Step 1 of 4

Number the equations

-3x - 3y = -33

... (1)

y = 3x - 9

... (2)

📖 Worked example

3 curry puffs and 1 packet of soya bean milk cost $4.70. 1 curry puff and 2 packets of soya bean milk cost $3.40. Find the cost of one curry puff and the cost of one packet of soya bean milk.

Step 1 of 6

Let one curry puff cost x dollars and one packet of soya bean milk cost y dollars. The first purchase

3x + y = 4.7

... (1)

Where the two equations come from	Equation
two lines meet at $P$	$P$ fits both lines' equations
opposite sides of a rectangle	$AB = DC$ , $AD = BC$
opposite angles of a parallelogram; co-interior angles	$\angle A = \angle C$ ; $\angle A + \angle B = 180^\circ$
a fraction $\dfrac{x}{y}$ that becomes $\dfrac{a}{b}$	cross-multiply: $b(\ldots) = a(\ldots)$

📖 Worked example

The lines $5x + 6y = 62$ and $-2x + y = -1$ meet at the point $P$ . Find the coordinates of $P$ .

Step 1 of 5

P is on both lines, so its x and y fit both equations: solve them together. Number the equations

5x + 6y = 62

... (1)

-2x + y = -1

... (2)

📖 Worked example

Step 1 of 6

Opposite sides of a rectangle are equal: AB = DC. Simplify

4x + 2y + 5 = 2x + 3y + 5

2x - y = 0

... (1)

📖 Worked example

Step 1 of 6

Write each fact as an equation, cross-multiply and simplify

\dfrac{x + 3}{y + 3} = \dfrac{7}{12}

12(x + 3) = 7(y + 3)

12x - 7y = -15

... (1)

\dfrac{x - 3}{y - 3} = \dfrac{1}{6}

6(x - 3) = 1(y - 3)

6x - y = 15

... (2)

Practise this →

Try a question: Linear equations, inequalities and simultaneous equations

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise linear equations, inequalities and simultaneous equations with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free