Solving Simple Equations
Goal: To learn how to solve simple or one‐step equaons.
An equaon is a mathemacal statement that says two expressions are equal.
Example: 4x+1=5, 2x+3=x-1
An expression is a mathemacal term or a sum or difference of mathemacal terms that may use numbers, variables, or both.
Examples: 2x+1,×3-2×2+5
To solve an equaon means to find the value of the variable that makes the statement true. This is also called finding the root of the equaon.
The opposite operaon of Subtraction
Mulplicaon Division
Squaring (x2) Square root()
Example 1: Solve x+7 = 15 by inspection
Solution: Since 8+7=15, then x=8
∴ x=8 is the root of x+7=15
Example 2: Solve 5n + 75 = 225 for n.
Soluon:
Method 1 ‐ Balanced Operaons
Noce that the equal signs
line up vercally and are in
the centre of the equaons
5n+75=225
5n+75-75=225-75 Subtract 75 from both sides of the equation
5n+0=150 Simplify
5n=150
5n5=1505 Divide by 5 both sides
n=30
∴ The solution to 5n+75=225 is n=30
Method 2 – Transposing
Noce that the equal signs
line up vercally and are in
the centre of the equaons
5n+75=225
5n=225-75 Move 75 from the left side to the right side
5n=150 Simplify
5n5=1505Divide by 5
n=30
Transposing is ONLY used for addion and subtracon.
Note that in both methods, we have isolated the variable on one side of the equaon (le side). To isolate the variable means to express an equaon so that the variable appears alone on one side of the equaon
To prove that we have the correct soluon (root) to an equaon, we perform a formal check by substung the root into the le and the right side of the equaon. Both sides must be equal.
Check the soluon to our last queson n = 30.
5n + 75 = 225
Le side of the equal sign (LS)
Right side of the equal sign (RS)
L.S =5n+75 Start with the original expressions.
=5(30)+75 Substute your answer into the variable(s).
=150+75 Simplify.
=225
Compare your results for each side.
R.S=225
∴ Since L.S=R.S, then n=30 is the correct root.
Example 3: Solve each equaon.
a) 2y = 14
2y2=142
∴y=7
b) x4=-3
4×4=4-3
∴x=-12
c) -t = 5
=-t-1=5-1
∴t=-5
Example 4: A high school football team has raised $1000 to spend on team jackets. The cost is $50 per jacket. Which equaon can be used to solve for the number of jackets the team can buy? Explain.
A. 50 = 1000n B. 50n = 1000 C. 1000n=50n
Let n represent the number of jackets
50n=1000
50n50=100050
n=20
∴ The team can buy 20 jackets.
OR
50 is the cost per jackets and 1000 is the money available
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