Ontario 10 Applied (MFM2P)
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Three step equations II
Lesson

Here we are looking at solving equations with fractions. This may look tricky, but we can follow the exact same steps that we did when solving equations with whole numbers.

Once again we will follow the reversed order of operations and make the variable the subject of the equation.

It may be helpful to write any fractions as improper fractions rather than mixed numbers.

Let's look through the process using an example: $2a+\frac{1}{3}=9$2a+13=9.

$2a+\frac{1}{3}$2a+13 $=$= $9$9 (Subtract $\frac{1}{3}$13 from both sides)
$2a$2a $=$= $8$8$\frac{2}{3}$23 (Convert the mixed fraction into an improper fraction)
$2a$2a $=$= $\frac{26}{3}$263 (Divide both sides by $2$2)
$a$a $=$= $\frac{26}{6}$266 (Simplify the answer)
$a$a $=$= $\frac{13}{3}$133  
Remember!

You can check your solution by substituting it back into the original equation.

Let' s do that now and substitute $a=\frac{13}{3}$a=133 into our equation:

$LHS$LHS $=$= $2a+\frac{1}{3}$2a+13
  $=$= $2\times\frac{13}{3}+\frac{1}{3}$2×133+13
  $=$= $\frac{26}{3}+\frac{1}{3}$263+13
  $=$= $\frac{27}{3}$273
  $=$= $9$9
  $=$= $RHS$RHS

 

Worked Examples

Question 1

Solve the following equation: $-x-\frac{7}{8}=3$x78=3

 
Question 2

Solve the equation $\frac{x}{-9}+6=6$x9+6=6

 
Question 3

Solve the equation $\frac{x-4}{-8}=-4$x48=4

 

Outcomes

10P.LR1.01

Solve first-degree equations involving one variable, including equations with fractional coefficients

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