Solve System of Linear Equations with Elimination and Substitution

 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.

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

1. Solve by elimination method:

x + y = 1

x – y = 3

2. Solve by substitution method:

   x + 2y = 2

-4x + 3y = 25

3. Solve by elimination method:

2x + 5y = -4

3x – y = 11

4. Solve by substitution method:

5x + 2y = 0

x – 3y = 0

5. Solve by elimination method:

x = 4y – 2

x = 6y + 8

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