SOLVING PAIRS OF LINEAR EQUATIONS BY SUBSTITUTION EXAMPLES

Solving Pairs of Linear Equations by Substitution Examples :

In this section, we will learn, how to solve linear equations by substitution method.

We use the following steps to solve a linear equations in substitution method.

Substitution Method

Step 1 :

Solve one of the equations for one of its variables.

Step 2 :

Substitute the expression from step 1 into the other equation and solve for the other variable.

Step 3 :

Substitute the value from step 2 into either original equation and solve to find the value of the variable in step 1.

Elimination Method

Step 1 :

By taking any one equations from the given two, first multiply by some suitable non-zero constant to make the co-efficient of one variable (either x or y) numerically equal.

Step 2 :

If both coefficients which are numerically equal of same sign, then we may eliminate them by subtracting those equations.

If they have different signs, then we may add both the equations and eliminate them.

Step 3 :

After eliminating one variable, we may get the value of one variable.

Step 4 :

The remaining variable is then found by substituting in any one of the given equations.

Solving Systems of Equations by Substitution Examples

Example 1 :

Solve the following pairs of linear equations by the elimination method and the substitution method

(i) x + y = 5 and 2 x – 3 y = 4

Solution :

x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Elimination method

(1) ⋅ 3 ==> 3x + 3y = 15

(1) + (2) 2x - 3y = 4

-------------------

5x = 19

x = 19/5

By applying x = 19/5 in (1) equation, we get

(19/5) + y = 5

y = 5 - (19/5)

y = (25 – 19)/5

y = 6/5

Therefore solution is x = 19/5 and y = 6/5

Now, let us do the same problem in substitution method :

Substitution method

x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Step 1 :

Find the value of one variable in terms of another variable

y = 5 – x

Step 2 :

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

2x – 3(5 – x) = 4

2x – 15 + 3 x = 4

5x = 4 – 15

5x = 19

x = 19/5

Step 3 :

Apply x = 19/5 in the equation y = 5 – x

y = 5 – (19/5)

y = (25 – 19)/5

y = 6/5

So, the solution is x = 19/5 and y = 6/5

Example 2 :

Solve the following pairs of linear equations by the elimination method and the substitution method

3x + 4y = 10 and 2x – 2y = 2

Solution :

3x + 4y = 10 ---------(1)

2x – 2y = 2 ---------(2)

Elimination method

(1) + (2) 3x + 4y = 10

(2) ⋅ 2 ==> 4 x - 4 y = 4

----------------

7x = 14

x = 14/7 = 2

By applying the value of x in (2), we get

2(2) – 2y = 2

4 – 2y = 2

-2y = 2 – 4

-2y = -2

y = 1

Therefore solution is x = 2 and y = 1

Substitution method

3 x + 4 y = 10 ---------(1)

2 x – 2 y = 2 ---------(2)

Step 1 :

Find the value of one variable in terms of another variable

-2y = 2 – 2 x

y = (2 x – 2)/2

y = x - 1

Step 2 :

Substitute this value of y in the other equation and reduce it to an equation in one variable.

3x + 4(x - 1) = 10

3x + 4x – 4 = 10

7x – 4 = 10

7x = 10 + 4

7x = 14

x = 2

Step 3 :

Apply x = 2 in the equation y = x - 1

y = 1

So, the solution is x = 2 and y = 1.

After having gone through the stuff given above, we hope that the students would have understood, solving systems of equations by elimination and substitution examples.

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