# Solve System of Linear Equations with Elimination and Substitution

Home | Online Math Tutoring | 8th Grade Math Tutoring | **Solve System of Linear Equations with Elimination and Substitution**

Let’s see this with the help of an example –

3*x* – 4*y* = 11 …Equation (1)

-3*x* + 2*y* = -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):

3*x* – 4(-2) = 11 3*x* + 8 = 11

3*x* = 11 – 8 3*x* = 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 3*x* – 4*y* = 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 -3*x* + 2*y* = -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.

*Example*: Solve the given system of equations by elimination method –

5*x* – 4*y* = 9 …Equation (1)

*x* – 2*y *= -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 –

5*x* – 4*y* = 9 …Equation (1) ——————— 5*x* – 4*y* = 9 …Equation (1)

*x* – 2*y *= -3 …Equation (2) ——————— -2*x* + 4*y *= 6 …Equation (2)

Adding the equations we get, 5x – 2x = 9 + 6

3*x* = 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 – 2*y* = -3

2*y* = 5 + 3

2*y* = 8 implies *y* = 4

**Check the solution:**

L.H.S of the equation 5*x* – 4*y* = 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* – 2*y *= -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*:

2*x* + *y* = 3 …Equation (1)

-5*x* + *y* = -4 …Equation (2)

** Step 1**: From equation (1),

*y*= 3 – 2

*x*

** Step 2 and 3**: Put

*y*= 3 – 2

*x*in equation (2)

*, –*5

*x*+ (3 – 2

*x*) = -4

-5x – 2x = -4 – 3

-7x = -7 ⇒ *x* = 1

** Step 4**: Put

*x*= 1 in equation (1),

*y*= 3 – 2

*x*

*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 2*x* + *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 -5*x* + *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* + 2*y* = 2

-4*x* + 3*y* = 25

- Solve by elimination method:

2*x* + 5*y* = -4

3*x* – *y* = 11

- Solve by substitution method:

5*x* + 2*y* = 0

*x* – 3*y* = 0

- Solve by elimination method:

*x* = 4*y *– 2

*x* = 6*y* + 8

##### Answer key

- (2, -1)
- (-4, 3)
- (3, -2)
- (0, 0)
- (-22, -5)

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