Solve System of Linear Equations with Elimination and Substitution
Grade 8 Math Worksheets
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Let’s see this with the help of an example –
3x – 4y = 11 …Equation (1)
-3x + 2y = -7 …Equation (2)
When we add the two equations, x-terms will be eliminated. This happens because the coefficients of the x-terms, 3 and -3, are opposites of each other.
So, adding we get:
Dividing both sides by -2, we get, y = -2
Now put y = -2 back in equation (1) or equation (2) to get the value of x. Let us put y = -2 in equation (1):
3x – 4(-2) = 11 3x + 8 = 11
3x = 11 – 8 3x = 3
x = 1
Check the solution: Put x = 1 and y = -2 in the given equations to check the answer.
Left Hand Side (L.H.S) of the equation 3x – 4y = 11:
3(1) – 4(-2) = 3 + 8 = 11, which is the same as right Hand Side (R.H.S) of the equation.
Since L.H.S = R.H.S, so the values are correct.
Similarly, L.H.S of the equation -3x + 2y = -7:
-3(1) + 2(-2) = -3 – 4 = -7, which is the same as R.H.S of the equation.
Since L.H.S = R.H.S, so the values are correct.
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Example: Solve the given system of equations by elimination method –
5x – 4y = 9 …Equation (1)
x – 2y = -3 …Equation (2)
Here we can either multiply equation (2) by 5 so that the coefficient of x in equation (2) is also 5 and then we can subtract both the equations to get the value of y, or we can multiply equation (2) by ‘-2’ and then add the two equations to get the value of x.
Here we multiply equation (2) by -2, and then add the equations –
5x – 4y = 9 …Equation (1) ——————— 5x – 4y = 9 …Equation (1)
x – 2y = -3 …Equation (2) ——————— -2x + 4y = 6 …Equation (2)
Adding the equations we get, 5x – 2x = 9 + 6
3x = 15
Dividing by 3 we get, x = 5
Put this value of x = 5 in any of the two equations to get the value of y, putting x = 5 in equation (2) we get:
5 – 2y = -3
2y = 5 + 3
2y = 8 implies y = 4
Check the solution:
L.H.S of the equation 5x – 4y = 9:
5(5) – 4(4) = 25 – 16 = 9, which is equal to the R.H.S of the equation.
L.H.S of the equation x – 2y = -3:
5 – 2(4) = 5 – 8 = -3, which is equal to the R.H.S of the equation.
Example: Solve the following equations for x and y:
2x + y = 3 …Equation (1)
-5x + y = -4 …Equation (2)
Step 1: From equation (1), y = 3 – 2x
Step 2 and 3: Put y = 3 – 2x in equation (2), –5x + (3 – 2x) = -4
-5x – 2x = -4 – 3
-7x = -7 ⇒ x = 1
Step 4: Put x = 1 in equation (1), y = 3 – 2x
y = 3 – 2(1)
y = 3 – 2
y = 1
Step 5: Check the solution: Put x = 1 and y = 1 in any of the given equations to check the answer.
Left Hand Side (L.H.S) of the equation 2x + y = 3:
2(1) + 1 = 2 + 1 = 3, which is equal to the R.H.S of the equation.
Since L.H.S = R.H.S so, the values are correct.
L.H.S of -5x + y = -4:
-5(1) + 1 = -5 + 1 = -4, which is equal to the R.H.S of the equation.
Since L.H.S = R.H.S so, the values are correct.
Check Point
- Solve by elimination method:
x + y = 1
x – y = 3
- Solve by substitution method:
x + 2y = 2
-4x + 3y = 25
- Solve by elimination method:
2x + 5y = -4
3x – y = 11
- Solve by substitution method:
5x + 2y = 0
x – 3y = 0
- Solve by elimination method:
x = 4y – 2
x = 6y + 8
Answer key
- (2, -1)
- (-4, 3)
- (3, -2)
- (0, 0)
- (-22, -5)
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