Inverse operations
To solve equations using algebra, the most important rule to remember is that if we apply operations to one side of the equation, we must also apply it to the other.
When applying operations to equations, we always apply the same step to both sides of the equation. This way, both sides of the equation will be equal once we solve the equation.
Making sure to follow this rule, we can isolate the variable in an equation by applying operations to both sides of the equation which reverse the operations applied to the variable.
Worked example
Example 1
Solve the equation: $\frac{2x+1}{5}=3$2x+15=3
Think: Consider what has happened to the starting value of $x$x. We can see that it is doubled, and then $1$1 is added to this. Finally it has been divided by $5$5. We want to reverse these operations in the reverse order. In other words, we want to undo the division by $5$5, then the addition of $1$1 and finally the multiplication by $2$2.
Do: Since multiplication reverses division, we can isolate $2x+1$2x+1 by multiplying both sides by $5$5.
$2x+1=15$2x+1=15
We can then reverse the addition of $1$1 by subtracting $1$1 from both sides.
$2x=14$2x=14
Finally, we can reverse the multiplication by $2$2, by dividing both sides by $2$2, giving us a final answer of $x=7$x=7.
Reflect: We can check our answer by substituting $x=7$x=7 into our original equation:
$\frac{2\times7+1}{5}=\frac{14+1}{5}$2×7+15=14+15
We can see that this simplifies to $3$3, so we know $x=7$x=7 is indeed the correct solution.
Equations with brackets
If we have an equation with one set of brackets such as $3\left(x-5\right)=9$3(x−5)=9 we can either expand the brackets before solving or, in this case as $3$3 is a factor of $9$9, divide both sides of the equation by $3$3. But in cases where we have two sets of brackets, we will first want to expand both sets of brackets before combining like terms. We can then solve the equation by performing inverse operations.
Practice question
Question 1
Solve the following equation:
$2\left(3x-5\right)+3\left(4x+6\right)=62$2(3x−5)+3(4x+6)=62
Enter each line of working as an equation.
Equations with fractions
Similarly, if we have an equation with a fraction such as $\frac{2x}{5}=4$2x5=4, we can just perform inverse operations to solve, but if we have an expression involving the addition or subtraction of two, or more, fractions we want to multiply both sides of the equation by the lowest common multiple of the fractions to eliminate them. We can then proceed as normal to solve.
Practice question
Question 2
Solve the following equation:
$\frac{5x}{6}-\frac{3x}{7}=-2$5x6−3x7=−2
Enter each line of working as an equation.
Equations with variables on both sides
To solve equations with variables on both sides of the equation, we want to use inverse operations (usually by adding or subtract terms) to eliminate the variables from one side of the equation. We can then combine like terms and solve using inverse operations.
Practice question
Question 3
Solve the following equation:
$3x+6=3\left(5x-4\right)+42$3x+6=3(5x−4)+42
Enter each line of working as an equation.