We reviewed the fundamental skills for solving equations in All Things Equation. Since then, we've learnt how to solve equations that also include decimals, fractions and brackets that need to be expanded.
Regardless of the operations involved in an equation, our aim is always the same. That is, to find the value of the unknown by 'getting it on its own'. So if decimals, fractions and brackets do appear in an equation, we solve these in just the same way, as you can see in the examples below.
Examples
Question 1
Solve: $x-1.39=8.67$x−1.39=8.67.
Think: We need to get $x$x by itself by adding 1.39 to both sides.
Do:
| $x$x | $=$= | $8.67+1.39$8.67+1.39 |
| $x$x | $=$= | $10.06$10.06 |
Question 2
Solve the following equation: $-x-\frac{7}{8}=3$−x−78=3.
Think: We need to get $x$x by itself, which we will do in a couple of steps.
Do:
| $-x-\frac{7}{8}$−x−78 | $=$= | $3$3 | (multiply each term by $8$8 to get rid of the denominator) |
| $-8x-7$−8x−7 | $=$= | $24$24 | (add $7$7 to both sides) |
| $-8x$−8x | $=$= | $31$31 | (divide both sides by $-8$−8) |
| $x$x | $=$= | $\frac{-31}{8}$−318 |
Question 3
Solve the following equation: $-7\left(x+1\right)=-28$−7(x+1)=−28.
Think: Again we aim to get $x$x by itself, which we will do in a couple of steps.
Do:
| $-7\left(x+1\right)$−7(x+1) | $=$= | $-28$−28 | (divide both sides by $-7$−7) |
| $x+1$x+1 | $=$= | $\frac{-28}{-7}$−28−7 | |
| $x+1$x+1 | $=$= | $4$4 | (subtract $1$1 from both sides) |
| $x$x | $=$= | $3$3 |
Question 4
Solve the following equation for $v$v:
$12v+6=110-v$12v+6=110−v
Question 5
Solve the following equation: $5x-\frac{104}{5}=x$5x−1045=x