While guessing and checking to find the solution to an equation does work, it isn't always the fastest method. Sometimes, we can interpret the equation as a simple calculation such as "what number divided by $4$4 equals $3$3?". But for harder questions involving multiple operations the answer will not always be obvious. In such cases, we can solve an equation using algebra.
Before we take a step forward to using algebra for solving equations, let's look back at what solving an equation requires.
In order to solve an equation containing a variable, we want to find a value that replaces the variable to make the equation true. While there are multiple methods available to us for solving equations in a systematic way, there are also times where we can just look at the equation and solve it.
For example, we can see quite easily that the equation $3+x=10$3+x=10 has the solution $x=7$x=7. We didn't use any particular method here, we just knew that $3+7$3+7 is equal to $10$10, so the number replacing the variable must be $7$7.
When solving equations like this, we are effectively reducing the equation to a 'fill in the blank' problem.
What is the solution to the equation $12-y=7$12−y=7?
Think: To solve the equation, we want to find a value for $y$y that makes the equation true. What value can we subtract from $12$12 to get $7$7?
Do: When solving the equation in this way, the problem becomes:
Fill in the blank: $12-\editable{}=7$12−=7
We know that $12-5$12−5 is equal to $7$7, so the solution to the equation is $y=5$y=5.
Reflect: When solving equations by looking at them, we think of the problem as a 'fill in the blank' type question and then solve it.
While this approach to solving equations is very easy and straightforward, it is only useful when the calculations are easy enough to solve just by looking at them. For more complex equations, we will need a method of simplifying the equation to one that we can solve more easily.
When trying to find the value for a variable that will solve the equation, we can avoid the guess and check method by isolating the variable on one side of the equation and directly solving for its value. This is also referred to as "making the variable the subject of the equation".
But how can we isolate the variable?
In order to isolate the variable, we want to apply operations to the equation so that the variable ends up alone on one side of the equation. In other words, we make it the subject of the equation.
What is the solution to the equation $w+8=14$w+8=14?
Think: We want to solve the equation by isolating $w$w in the equation. In order to do this, we will need to remove the $8$8 that has been added to $w$w.
Do: Since subtraction reverses addition, we can solve for $w$w by subtracting $8$8 from both sides of the equation, as this will cancel out the $8$8 that is with the $w$w since $8-8=0$8−8=0.
This gives us:
$w+8-8=14-8$w+8−8=14−8
Notice that if we add $8$8 to $w$w, then subtract $8$8 from $w$w, the addition and subtraction will cancel out and we will be left with only $w$w on the left-hand side of the equation:
$w=14-8$w=14−8
We can then evaluate the subtraction to find the solution to the equation:
$w=6$w=6
Reflect: To isolate $w$w, we applied an operation that would reverse the addition of $8$8. Once $w$w was isolated, we evaluated the other side of the equation to find the solution.
Notice that when we applied the subtraction of $8$8 to the equation, we applied it to both sides of the equation. This is very important!
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.
Consider the equation $9p=36$9p=36.
Which of the following options will make $p$p the subject of the equation?
Divide both sides by $9$9
Subtract $9$9 from both sides
Multiply both sides by $0$0
Multiply both sides by $9$9
Apply the operation from part (a) to the equation to find the solution.
Enter each step as an equation.
When isolating the variable, we need to know which operation to apply so that the variable will be on its own at the end. The fastest way to do this is to reverse any operations that are currently being applied to the variable.
To do this, we need to know which operations are reversed by which.
Operation | Reverse operation | Example |
---|---|---|
Addition Subtraction |
Subtraction Addition |
$x+4-4=x$x+4−4=x |
Multiplication Division |
Division Multiplication |
$y\times4\div4=y$y×4÷4=y |
We can see that four basic operations can be sorted into pairs,
As we can see in the table, when operations that reverse each other are applied to the same variable (or number) they will cancel out.
What is the reverse of 'adding $7$7'?
Think: '"Adding $7$7" implies addition, so we are looking for the reverse operation of addition.
Do: Since subtraction reverses addition and the amount we subtract should be equal to the amount added, the reverse is 'subtracting $7$7'.
What is the reverse of 'subtracting $13$13'?
Think: 'Subtracting $13$13' implies subtraction, so we are looking for the reverse operation of subtraction.
Do: Since addition reverses subtraction ($-13+13=0$−13+13=0) and the amount we add should be equal to the amount subtracted, the reverse is 'adding $13$13'.
What is the reverse of 'multiplying by $4$4'?
Think: 'Multiplying by $4$4' implies multiplication, so we are looking for the reverse operation of multiplication.
Do: Since division reverses multiplication and the amount we divide by should be equal to the amount multiplied by, the reverse is 'dividing by $4$4'.
What is the reverse of 'dividing by $10$10'?
Think: 'Dividing by $10$10' implies division, so we are looking for the reverse operation of division.
Do: Since multiplication reverses division and the amount we multiply by should be equal to the amount divided by, the reverse is 'multiplying by $10$10'.
Reflect: When finding the reverse operation, the operation will always be the reverse of that in the original operation while the number stays the same.
Remember that terms like $4n$4n represent the multiplication $n\times4$n×4 and fractions like $\frac{m}{6}$m6 represent the division $m\div6$m÷6.
Solve the equation $v-4=44$v−4=44.
Enter each step as an equation.
When we reverse an operation in an equation, we are applying the reverse operation to both sides of the equation.
Consider the equation $\frac{x}{7}=12$x7=12.
We can write this using the basic operations as $x\div7=12$x÷7=12
To isolate $x$x, we want to reverse the division by $7$7. Since multiplication reverses division, the reverse operation will be 'multiply by $7$7'. Applying this to both sides of the equation gives us:
$x\div7\times7=12\times7$x÷7×7=12×7
After cancelling out the division and multiplication of $7$7, we get:
$x=12\times7$x=12×7
Notice that when we applied the reverse operation, it had the same effect as turning division on the left-hand side into multiplication on the right-hand side.
This shows us that, by changing both the side and sign of an operation, we can reverse the operation in a single step.
Solve the equation $24+p=59$24+p=59.
Enter each step as an equation.