CONNECT: Algebra
SOLVING EQUATIONS
An Algebraic equation is where two algebraic expressions are equal to each
other. Finding the solution of the equation means finding the value(s) of the
variable which makes the equation true. There may be one or more, or even
no solutions to the equation.
(An example of an equation which has no solution using the set of Real
numbers is 𝑥𝑥 = √−9. Try this on your calculator. We can’t find the square
root of a negative number because no matter what real number we square we
always end up with a positive result.)
One of the easiest ways in which to solve an equation is to use the idea of
backtracking. With this, you start by working forwards, working out what has
been done to the variable step by step, then working backwards to undo that
procedure. Here is an example:
Find the value of 𝑥𝑥 if 5𝑥𝑥 + 7 = 22.
Firstly, we map what has been done to the variable (in this case, 𝑥𝑥 ):
x5
𝑥𝑥
+7
5𝑥𝑥
5𝑥𝑥
22
Now we work out what to do to go backwards, starting from the righthand box.
x5
𝑥𝑥
+7
5𝑥𝑥
÷5
1
5𝑥𝑥
–7
22
While we are doing this, we can also fill in the numbers in the boxes:
x5
+7
𝑥𝑥
5𝑥𝑥
3
15
÷5
5𝑥𝑥
–7
22
Last, check that the value of 𝑥𝑥 fits the original equation:
𝑥𝑥 = 3 gives 5𝑥𝑥 + 7 = 5 x 3 + 7 which is 22 and agrees with the original
equation.
We can also solve this equation with the “Balance both sides” technique,
(which is a different version of the backtracking method):
5𝑥𝑥 + 7 = 22
Take 7 away from both sides of the equals sign (undo the +7):
5𝑥𝑥 + 7 − 7 = 22 − 7.
So, 5𝑥𝑥 = 15 .
backtracking.)
(Note, this is the same as the middle boxes in the
Divide both sides by 5 (undo the x5):
5𝑥
5
=
𝑥𝑥 = 3
15
5
Try the backtracking technique on a more difficult equation:
7𝑥−2
10
2
=4
x7
𝑥𝑥
–2
7𝑥𝑥
÷10
7𝑥𝑥 − 2
10
7𝑥𝑥 − 2
÷7
+2
x10
4
Now the balancing both sides technique:
7𝑥−2
=4
10
Multiply both sides by 10 (to undo the ÷10)
7𝑥𝑥−2
( 10 ) × 10 = 4 × 10
7𝑥𝑥 − 2 = 40
Add 2 to both sides (to undo the –2)
7𝑥𝑥 − 2 + 2 = 40 + 2
7𝑥𝑥 = 42
Divide both sides by 7 (to undo the x7)
7𝑥
7
=
42
𝑥𝑥 = 6
7
Check 𝑥𝑥 = 6 works in the original equation:
I will use LHS, RHS notation here. We want to make sure that the left hand
side of the equals sign (LHS) of the original equation is the same as the right
hand side (RHS).
LHS
3
=
7𝑥−2
=
7×6−2
10
10
RHS = 4
=
=
42−2
10
40
10
= 4 = RHS
So 𝑥𝑥 = 6 is the correct solution to the equation.
A last example:
Solve
(4𝑥−7)×5+3
2
=4
Using backtracking – you can fill in the blanks and work backwards:
x4
𝑥𝑥
–7
4𝑥𝑥
x5
4𝑥𝑥 − 7
+3
(4𝑥𝑥 − 7) × 5
÷2
(4𝑥𝑥 − 7) × 5+3
(4𝑥𝑥 − 7) × 5 + 3
2
).2 = 𝑥𝑥 deniatbo evah dluohs uoY(
The idea with solving equations using either technique is to work out what has
been done to the variable (often 𝑥𝑥 ) and then to undo those steps in the
reverse order.
Here are some for you to try: You can check your solutions with the worked
solutions at the end of this resource.
Solve each of the following equations using either the backtracking method or
the balancing both sides technique. Check your answer using the LHS/RHS
method.
4
4
1. (2x – 1) x 3 = 21
2. 5(x – 2) –3 = 42
3.
6𝑥+8
4.
11(3𝑥+5)
4
2
= 11
= 44
If you need help with any of the Maths covered in this resource (or any other
Maths topics), you can make an appointment with Learning Development
through Reception: phone (02) 4221 3977, or Level 3 (top floor), Building 11,
or through your campus.
Unfortunately, the Backtracking method is only valid for equations when the
variable is on one side of the equation only. However, the general techniques
of “undoing” each step work in every situation.
Before we go on, check that you know the inverse procedure (the “undoing”
technique) for each of the following operations:
OPERATION
INVERSE OPERATION
+
x
squaring
Solve √𝑥𝑥 + 1 = 3
What has been done to the variable (x) here?
5
Firstly, 1 has been added, then the square root has been taken of the result.
We need to undo these operations in the reverse order.
So, firstly, square both sides to undo the √
(√𝑥𝑥 + 1)2 = 32
𝑥𝑥 + 1 = 9
Take away 1 from both sides
𝑥𝑥 + 1 – 1 = 9 – 1
𝑥𝑥 = 8
Check solution.
LHS = √𝑥𝑥 + 1
RHS = 3
= √8 + 1
= √9
= 3 = RHS
So 𝑥𝑥 = 8 is the correct solution.
6
Examples of equations where the variable is on both sides
Solve: 5x – 6 = 3x + 2
The idea is to move the variables together. For that to happen in this example,
we could take 3x away from both sides:
5x – 6 – 3x = 3x + 2 – 3x
2x – 6 = 2
Now proceed as above, either using backtracking or balancing both sides (I
will use balancing both sides because it is easier to type!)
2x – 6 = 2
Add 6 to both sides:
2x = 8
x=4
Check the solution. This time you need to substitute into both LHS and RHS.
LHS = 5x – 6
RHS = 3x + 2
=5x4–6
=3x4+2
= 20 – 6
= 12 + 2
= 14
= 14
So x = 4 is the correct solution.
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Further examples of equations where the variable is on both sides
Example 1. Solve: 4x + 3 = 8x – 5
This time, we could take 4x away from both sides.
4x + 3 – 4x = 8x – 5 – 4x
3 = 4x – 5
Now, add 5 to both sides.
3 + 5 = 4x – 5 + 5
8 = 4x
2=x
This is the same as x = 2.
Check the solution.
LHS = 4x + 3
RHS = 8x – 5
=4x2+3
=8x2–5
=8+3
= 16 – 5
= 11
= 11
So x = 2 is the correct solution.
Example 2. Solve 5(2𝑥𝑥 + 1) = 3�4𝑥𝑥 – 1�.
Here it is easiest to expand both sides (multiply the brackets)
10𝑥𝑥 + 5 = 12𝑥𝑥 – 3
Take 10𝑥𝑥 away from both sides.
8
10𝑥𝑥 + 5 – 10𝑥𝑥 = 12𝑥𝑥 – 3 – 10𝑥𝑥
5 = 2𝑥𝑥 – 3
Add 3 to both sides
5 + 3 = 2𝑥𝑥 – 3 + 3
8 = 2𝑥𝑥
Divide both sides by 2
8 2𝑥𝑥
=
2
2
So 𝑥𝑥 = 4
4 = 𝑥𝑥
Check the solution.
LHS = 5(2𝑥𝑥 + 1)
= 5(2 x 4 + 1)
RHS = 3�4𝑥𝑥 – 1�
= 3(4 x 4 – 1)
= 5(9)
= 3(15)
= 45
= 45
So 𝑥𝑥 = 4 is the correct solution.
If you need help with any of the Maths covered in this resource (or any other
Maths topics), you can make an appointment with Learning Development
through Reception: phone (02) 4221 3977, or Level 3 (top floor), Building 11,
or through your campus.
Here are some for you to try. Solve each equation for 𝑥𝑥 :
1.
9
3𝑥𝑥 + 5 = 5𝑥𝑥 + 1
2. 4𝑥𝑥 – 1 = 3(2𝑥𝑥 – 5)
3. 12 – 2𝑥𝑥 = 𝑥𝑥 – 3
WORKED SOLUTIONS TO EQUATIONS
1. (2x – 1) x 3 = 21
Backtracking method:
x2
–1
𝑥𝑥
x3
2𝑥𝑥
4
2𝑥𝑥 − 1
8
÷2
7
+1
Balancing both sides technique:
(2x – 1) x 3 = 21
Divide both sides by 3:
(2𝑥−1)×3
3
=
2x – 1 = 7
21
3
Add 1 to both sides:
2x – 1 + 1 = 7 + 1
2x = 8
Divide both sides by 2:
2𝑥
2
8
=2
𝑥𝑥 = 4
Check solution:
LHS = (2x – 1) x 3
= (2 x 4 – 1) x 3
= (8 – 1) x 3
=7x3
= 21 = RHS
So 𝑥𝑥 = 4 is the correct solution.
10
(2𝑥𝑥 − 1) × 3
RHS = 21
÷3
21
2. Solve 5(x – 2) –3 =42
Backtracking method:
–2
x5
𝑥𝑥
𝑥𝑥 − 2
11
9
+2
5(𝑥𝑥 − 2)
÷5
Balancing both sides technique:
5(x – 2) –3 = 42
Add 3 to both sides:
5(x – 2) –3 + 3 = 42 + 3
5(𝑥𝑥 – 2) = 45
Divide both sides by 5:
5(𝑥−2)
5
=
x–2=9
45
5
Add 2 to both sides
x–2+2=9+2
x = 11
Check solution:
LHS = 5(x – 2) –3
= 5(11 – 2) – 3
11
–3
RHS = 42
45
(2𝑥𝑥 − 1) × 3
+3
42
=5x9–3
= 45 – 3
= 42 = RHS
So x = 11 is the correct solution.
3.
6𝑥+8
4
= 11
Backtracking method:
x6
𝑥𝑥
+8
6𝑥𝑥
6
÷6
36
6𝑥𝑥 + 8
–8
Balancing both sides technique:
6𝑥+8
4
= 11
Multiply both sides by 4:
6𝑥+8
4
× 4 = 11 × 4
6𝑥𝑥 + 8 = 44
Take 8 away from both sides:
6𝑥𝑥 + 8 – 8 = 44 – 8
6𝑥𝑥 = 36
Divide both sides by 6:
𝑥𝑥 = 6
12
÷4
44
x4
6𝑥𝑥 + 8
4
11
Check solution:
LHS =
6𝑥+8
=
=
RHS = 11
4
6×6+8
4
44
4
= 11= RHS
So 𝑥𝑥 = 6 is the correct solution.
4.
11(3𝑥+5)
= 44
2
Backtracking method:
x3
𝑥𝑥
1
+5
3𝑥𝑥
÷3
3
x11
–5
Balancing both sides technique:
11(3𝑥𝑥 + 5)
= 44
2
11(3𝑥+5)
2
Multiply both sides by 2:
× 2 = 44 × 2
11(3𝑥𝑥 + 5) = 88
13
11(3𝑥𝑥 + 5)
3𝑥𝑥 + 5
8
÷2
÷11
88
x2
11(3𝑥𝑥 + 5)
2
44
Divide both sides by 11:
11(3𝑥+5)
11
=
88
11
3𝑥𝑥 + 5 = 8
Take 5 away from both sides:
3𝑥𝑥 + 5– 5 = 8– 5
3𝑥𝑥 = 3
Divide both sides by 3:
3𝑥𝑥 3
=
3
3
𝑥𝑥 = 1
Check solution:
LHS =
=
=
=
=
11(3𝑥+5)
2
11(3×1+5)
RHS = 44
2
11(3+5)
2
11×8
2
88
2
= 44 = RHS
So
𝑥𝑥 = 1 is the correct solution.
14
From page 9:
3𝑥𝑥 + 5 = 5𝑥𝑥 + 1
1.
Take 3𝑥𝑥 from both sides:
3𝑥𝑥 + 5 – 3𝑥𝑥 = 5𝑥𝑥 + 1– 3𝑥𝑥
5 = 2𝑥𝑥 + 1
Take 1 from both sides:
5 – 1 = 2𝑥𝑥 + 1 – 1
4 = 2𝑥𝑥
Divide both sides by 2:
2 = 𝑥𝑥
This is the same as 𝑥𝑥 = 2.
Check the solution:
LHS = 3𝑥𝑥 + 5
=3x2+5
= 11
RHS = 5𝑥𝑥 + 1
=5x2+1
= 11
∴ 𝑥𝑥 = 2
2. 4𝑥𝑥 – 1 = 3(2𝑥𝑥 – 5)
Expand RHS:
4𝑥𝑥 – 1 = 6𝑥𝑥 – 15
Take 4𝑥𝑥 from both sides:
– 1 = 2𝑥𝑥 – 15
Add 15 to both sides:
14 = 2𝑥𝑥
Divide both sides by 2 and reverse the equation:
𝑥𝑥 = 7.
15
Check solution:
LHS = 4𝑥𝑥 – 1
=4x7–1
= 27
RHS = 3(2𝑥𝑥 – 5)
= 3(2 x 7 – 5)
= 3(9)
= 27
∴ 𝑥𝑥 = 7 is the solution.
3. 12 – 2𝑥𝑥 = 𝑥𝑥 – 3
Add 2𝑥𝑥 to both sides:
12 = 3𝑥𝑥 – 3
Add 3 to both sides:
15 = 3𝑥𝑥
Divide both sides by 3 and reverse equation:
𝑥𝑥 = 5
Check solution:
LHS = 12 – 2 x 5
=2
RHS = 5 – 3
=2
∴ 𝑥𝑥 = 5 is the correct solution.
16