Equations
An equation is a statement that connects two expressions with the equals sign (=) to express their equality.
Solving One-variable Equations Using Different Operations
For some values, equations may be true, while for others, they may be false. A value that makes an equation true is called a solution.
The table below shows the formation of the correct solution:
Hence, when the value of a = 6, it will form a unique solution for the equation a + 2 = 8.
Opposite operations are also referred to as inverse operations. They are operations that counteract the effects of other operations. Addition, for example, is the inverse of subtraction, and multiplication is the inverse of division.
Properties
The addition property of equality
The two sides of an equation remain equal when the same number is added to each side.
- Algebraic application:
a – 2 = 4
+ 2 +2
a = 6
- Numerical application:
6 = 6
+4 +4
10 = 10
The subtraction property of equality
The two sides of an equation remain equal when the same number is subtracted from each side.
- Algebraic application:
a + 2 = 4
– 2 – 2
a = 2
- Numerical application:
6 = 6
– 4 – 4
2 = 2
The multiplication property of equality
When both sides of an equation are multiplied by the same non-zero number, the two sides remain equal.
- Algebraic application:
\(\frac{a}{2}=4\)
\(\frac{a}{2}.2=4.2\)
a = 8
- Numerical application:
\(\frac{6}{2}=3\)
\(\frac{6}{2}.2=3.2\)
6 = 6
The multiplicative inverse property
The non-zero number a and its reciprocal, 1/a, have a product of one.
- Algebraic application:
\(n.\frac{1}{n}=\frac{1}{n}.n=1\), n ≠ 0
- Numerical application:
\(6.\frac{1}{6}=1\)
The division property of equality
When both sides of an equation are divided by the same non-zero number, the two sides remain equal.
- Algebraic application:
2x = 4
\(\frac{2X}{2}=\frac{4}{2}\)
x = 2
- Numerical application:
\(6.2=12\)
\(6.\frac{2}{2}=\frac{12}{2}\)
6 = 6
Solved Examples
Example 1: Solve the following equation to check whether the equation is forming a solution or not.
(a) 3a=42,a=14
(b) x-7=22,x=30
Solution:
Part (a)
We have: 3a = 42, a = 14
3(14) = 42 Substitute a with 14
42 = 42
It will form a unique solution.
Part (b)
We have: x – 7 = 22, x = 30
30 – 7 = 22 Substitute x with 30
23 = 22
It will not form a unique solution.
Example 2: Solve the following equations using different operations.
(a) 15 = x – 3
(b) a + 3 = 8
(c) \(\frac{a}{2}\) = 6
(d) 35 = 5x
Solution:
Part (a)
We have: 15 = x – 3
15 = x – 3
+3 + 3 Using the addition property of equality
18 = x
The solution is x = 18.
Part (b)
We have: a + 3 = 8
a + 3 = 8
-3 – 3 Using the subtraction property of equality
a = 5
The solution is a = 5.
Part (c)
We have: \(\frac{a}{2}=6\)
\(\frac{a}{2}=6\)
.2 .2 Using the multiplication property of equality
a = 12
The solution is a = 12.
Part (d)
We have: 35 = 5x
35/5 = 5x/5
Using the division property of equality
7 = x
The solution is x = 7.
Example 3: Your uncle gives you $60 to help you purchase the mobile phone in the picture. You have $25.50 left after purchasing the phone. How much money did you have before your uncle gave you $50?
Solution:
Write an equation using the word sentence of the given question:
Amount left ($) = Initial Amount ($) + Uncle’s Money ($) – Price of the phone ($)
Let the initial amount be i and form an equation:
= 25.50 = i + 60 – 275.95
= 25.50 + 275.95 = i + 60 – 275.95 + 275.95 Using the addition property of equality
= 301.45 = i + 60
= 301.45 – 60 = i + 60 -60 Using the subtraction property of equality
= 241.45 = i
The total amount of money you have is $241.45.
Example 4: In a park, a rectangular LED “screen” covers 7500 square metres. The screen has a 60-metre width. What is the length of the screen?
Solution:
In the formula for the area of a rectangle, A = lw, substitute values for the area A and the width. Calculate the length l.
The area of a rectangle = A = lw
Substitute the value for A = 7500, and w = 60
7500 = \(60l\)
7500 = \(60l\) Using the division property of equality
\(\div 60 \div 60\)
\(l=125\)
Therefore, the screen in the park is 125 meters long.
Here are the steps for solving linear equations with one variable:
Step 1: Using LCM, remove any fractions that may exist.
Step 2: Make both sides of the equation as simple as possible.
Step 3: Identify and isolate the variable.
Step 4: Double-check your answer
When solving multi-variable, multi-step equations, the first rule is to make sure you have the same number of equations as the number of variables in the equations.
Then, for one of the variables, solve one of the equations and plug that expression into the other equation for what it equals.
A multivariable function is simply a function with multiple numbers as the input or output.
A single-variable function, on the other hand, has a single-number input and a single-number output.