Solve Simultaneous Equations By Elimination

Before viewing this page, it would be helpful to learn how to Solve Simultaneous Equations By Graphing.

The purpose of solving simultaneous equations is to find the same x-value and the same y-value that satisfies both equations.

To solve, the like terms of two equations are lined up under each other, the whole equations are added or subtracted, and one variable is eliminated in the process.

When adding numbers, we line up the digits in columns of hundreds, tens and units...

1 2 3
+ 4 5 6
5 7 9

When adding whole equations, we line up the like or similar terms, but a variable is eliminated.
This is how the process starts...

3x + 2y + 5 = 0
+ (4x – 2y + 9 = 0)
7x + 14 = 0

Here's a demonstration...


Example One - Elimination by Adding

Solve these equations by elimination:
4x + y = 9
6x – y = 11

Answer:

4x + y = 9
+ (6x – y = 11)
10x = 20

10x = 20
x = 2

Now, we substitute the x value into one of the equations to find the y value...
4x + y = 9
8 + y = 9
y = 1

The simultaneous solution is x = 2 and y = 1.


Example Two - Elimination by Subtracting

Solve these equations by elimination:
5y + 7x = 8
3y + 7x = 2

Answer:
Remember that subtracting the second equation means that all signs of the second equation change to their opposite signs.

5y + 7x = 8
– (3y + 7x = 2)
2y = 6

2y = 6
y = 3

Now, we substitute the y value into one of the equations to find the x value...
5y + 7x = 8
15 + 7x = 8
7x = 8 – 15
7x = –7
x = –1

The simultaneous solution is x = –1 and y = 3.


Example Three - Lining Up Whole Equations First

Solve these equations by elimination:
4x – 3y = 26
x – 11 = 3y

Answer:

4x – 3y = 26
– (x – 3y = 11)
3x = 15

3x = 15
x = 5

Now, we substitute the x value into one of the equations to find the y value...
x – 11 = 3y
5 – 11 = 3y
–6 = 3y
–2 = y
y = –2

The simultaneous solution is x = 5 and y = –2.


Example Four - Multiplying Whole Equations First

Solve these equations by elimination:
4x + 3y = 11
3x – 2y = 21

Answer:
I intend to eliminate the y variable.
I will multiply the first equation by 2.
I will multiply the second equation by 3.
Then there will be 6y in both equations...

(4x + 3y = 11) × 2
(3x – 2y = 21) × 3

8x + 6y = 22
9x – 6y = 63

Now, I will add these equations to eliminate y...

8x + 6y = 22
+ (9x – 6y = 63)
17x = 85

17x = 85
x = 5

Now, we substitute the x value into any equation to find the y value...
4x + 3y = 11
20 + 3y = 11
3y = 11 – 20
3y = –9
y = –3

The simultaneous solution is x = 5 and y = –3.

Questions - Solve by Elimination

Q1.
x + y = 6
x – y = 4

Q2.
3x + 5y = 14
2x – 5y – 1 = 0

Q3.
x – 2y = –5
2x + 3y – 4 = 0

Answers
A1. x = 5, y = 1
A2. x = 3, y = 1
A3. x = –1, y = 2