Solving Linear Equations (Cambridge International IGCSE Mathematics)

Solving linear equations

What are linear equations?

• A linear equation is one that can be written in the form $ax+b=c$

  • $a,b,$ and $c$* *are numbers and $x$is the variable

    • 2x + 3 = 5
    • 3x + 4 = 1
    • x - 5 = -3

• The greatest power of x is 1

  • There are no terms like x²

How do I solve linear equations?

• You need to use operations like adding, subtracting, multiplying and dividing to get x on its own

• Any operation you do to one side of the equation must also be done to the other side

• For example, to solve $2x+1=9$ look at the +1 on the left

  • Undo this by subtracting 1 from both sides and simplifying

$\begin{gathered} 2 x + 1 & = 9 \end{gathered}$

$\begin{gathered} ( - 1 ) \end{gathered} \begin{gathered} ( - 1 ) \end{gathered}$

$\begin{gathered} 2 x & = 8 \end{gathered}$

• This equation is now easier to solve
• 2x is 2 × x so undo this by dividing both sides by 2 and simplifying

$\begin{gathered} 2 x & = 8 \end{gathered}$

$\begin{gathered} ( \div 2 ) \end{gathered} \begin{gathered} ( \div 2 ) \end{gathered}$

$\begin{gathered} x = 4 \end{gathered}$

• The solution to the equation is x = 4

• Adding 1 was undone by subtracting 1

• Multiplying by 2 was undone by dividing by 2

  • Addition and subtraction are said to be inverse(opposite) operations
  • Multiplication and division are also inverse operations

Does the order of steps matter?

• As long as each step is applied correctly, the order in which inverse operations are applied does not matter

  • Applying the operations in one order may be easier than another

• Consider $4x+8=12$

  • It is easier to first subtract 8 from both sides

$4x=4$

• Then divide both sides by 4

$\begin{gathered} x = 1 \end{gathered}$

• If you want to first divide by 4, a common mistake is to write $x+8=3$

  • This is incorrect as 8 has not been divided by 4
  • You must divide every term by 4

$\begin{gathered} \frac{4 x}{4} + \frac{8}{4} & = \frac{12}{4} \\ x + 2 & = 3 \end{gathered}$

• Then subtract 2 from both sides

$\begin{gathered} x = 1 \end{gathered}$

How do I solve linear equations with negative numbers?

• For example, $2-3x=10$

  • Subtract 2 from both sides and simplify

$\begin{gathered} 2 - 3 x & = 10 \end{gathered}$

$\begin{gathered} ( - 2 ) \end{gathered} \begin{gathered} ( - 2 ) \end{gathered}$

$\begin{gathered} - 3 x & = 8 \end{gathered}$

• Then divide both sides by -3 and simplify

$\begin{gathered} - 3 x & = 8 \end{gathered}$

$\begin{array}{rcl} ( \div - 3 ) \end{array} \begin{array}{rcl} ( \div - 3 ) \end{array}$

$\begin{gathered} x = - \frac{8}{3} \end{gathered}$

• Some people prefer to write $2-3x=10$ as $-3x+2=10$, which is equivalent

  • You then subtract 2 and divide by -3 as before

Exam Hint Substitute your answer back into the original equation to check you got it right!

Worked Example Solve the equation

$9-7x=5$

Answer:

Subtract 9 from both sides of the equation

$\begin{gathered} - 7 x & = - 4 \end{gathered}$

Divide both sides by -7 Remember that a negative divided by a negative will result in a positive number

$x=\frac{-4}{-7}$

$x=\frac{4}{7}$

How do I solve linear equations with brackets?

• If a linear equation involves brackets, expand the brackets first

• For example, solve $2 ( x - 3 ) = 10$

  • Expand the brackets

$2x-6=10$

• Then solve as shown previously

  • Add 6 then divide by 2

$\begin{gathered} 2 x & = 16 \\ x = 8 \end{gathered}$

• Expanding brackets first will always work, but you can also divide first

  • Dividing both sides of $2 ( x - 3 ) = 10$ by 2 gives $( x - 3 ) = 5$

    • which gives $x=8$

  • This method works but can lead to harder fractions

How do I solve linear equations with fractions?

• If a linear equation contains fractions, multiply both sides by the lowest common denominator

• For example, $\frac{x}{5}+4=\frac{9}{2}$

  • The lowest common denominator of 5 and 2 is 10
  • Multiply all terms on both sides by 10

$\begin{gathered} ( 10 \times \frac{x}{5} ) + ( 10 \times 4 ) & = 10 \times \frac{9}{2} \\ \frac{10 x}{5} + 40 & = \frac{90}{2} \end{gathered}$

• Simplify the fractions

$\begin{gathered} 2 x + 40 & = 45 \end{gathered}$

• Now solve as before, by subtracting 40, then dividing by 2

$\begin{gathered} 2 x & = 5 \\ x = \frac{5}{2} \end{gathered}$

• Unless the question specifies otherwise, you can leave the answer like this

  • A decimal or mixed number would also be accepted

What if the unknown is on the denominator?

• For example $\frac{4}{x-2}=3$

  • Multiply both sides of the equation by the denominator

$\frac{4}{x - 2} \times ( x - 2 ) = 3 ( x - 2 )$

• Simplify the fractions, and expand any brackets

$4 = 3 ( x - 2 ) 4 = 3 x - 6$

• Now solve as before, by adding 6 to both sides, then dividing by 3

$\begin{gathered} 10 = 3 x \\ \frac{10}{3} & = x \end{gathered}$

Worked Example (a) Solve the equation

$5 ( 3 - 4 x ) + 1 = 26$

Answer:

Expand the bracket

$\begin{gathered} 15 - 20 x + 1 & = 26 \end{gathered}$

Simplify

$\begin{gathered} 16 - 20 x & = 26 \end{gathered}$

It can be helpful to consider 16 - 20x as -20x + 16 Subtract 16 from both sides

$\begin{gathered} - 20 x & = 10 \end{gathered}$

Divide both sides by -20 and simplify

$\begin{gathered} x = \frac{10}{- 20} \\ x = - 1 \frac{1}{2} \end{gathered}$

$x=-\frac{1}{2}$

-0.5 is also accepted

(b) Solve the equation

$\frac{5x}{4}=\frac{1}{2}$

Answer:

The lowest common denominator of 4 and 2 is 4 Multiply both sides by 4

$\begin{gathered} 4 \times \frac{5 x}{4} & = 4 \times 1 \frac{1}{2} \end{gathered}$

Simplify (cancel) the fractions

$\begin{gathered} 5 x & = 2 \end{gathered}$

To solve this equation, divide both sides by 5

$x=\frac{2}{5}$

How do I solve linear equations with x terms on both sides?

• Collect the x terms (or whichever variable is involved) together on one side

  • To do this, remove all the x terms from one side

    • It is easiest to remove the smallest x term to avoid negatives

• For example, $4x-7=11+x$

• Remove the x term on the right-hand side, by subtracting x from both sides

$\begin{gathered} 4 x - 7 & = 11 + x \end{gathered}$

$\begin{gathered} ( - x ) ( - x ) \end{gathered}$

$\begin{gathered} 3 x - 7 & = 11 \end{gathered}$

• There are no longer any x terms on the right

• This now has the same form as previously seen

  • Solve by adding 7 then dividing by 3

$\begin{gathered} 3 x - 7 & = 11 \\ 3 x & = 18 \\ x = 6 \end{gathered}$

Worked Example Solve the equation

$4-5x=6x-29$

Answer:

Remove the x terms from either side We will remove them from the left as -5**x is smaller than 6x

Add 5**x to both sides

$\begin{gathered} 4 = 11 x - 29 \end{gathered}$

Get 11 x on its own by adding 29 to both sides

$33=11x$

*Divide both sides by 11 to find *x

$3=x$

$x=3$