A System Equations Solved by Algebraic Methods

Solution by Substitution

This method is easily applied when at least one equation gives the value of one unknown of the other.

To solve a system of Linear Equation in two variables:

1. Use either of the equations to solve for one variables.(Always select the equation where in the variable is easier to work with).

2. Substitute the expression obtained in step 1 in the other equation.

3. Solve the resulting equation to find the value of the variable.

4. Substitute the value of this variables in the simpler equation to find the value of the other variable.

5. The resulting no pair is the solution.

6. Check the results in the two original equations.

Examples:
A. Solve the system:
3x + 2y = 11 (1)
y = 4x (2)

Solution:
3x + 2y = 11
y = 4x

Substitute 4x for y in equation (1)

3x + 2(4x) = 11
3x + 8x = 11
11x = 11
x = 11/11 or 1

Substitute 1 for x in equation (2)

y = 4(1)
y = 4

Check:
3x + 2y = 11
3(1)+ 2(4) = 11
11 = 11 /

y = 4x
4 = 4(1)
4 = 4 /

The solution is the ordered pair (1,4).

B. Solve the system:
5x + 3y = 3 (1)
y = 8 - 4x (2)


Solution:
5x + 3y = 3
y = 8 - 4x

Substitute 8 - 4x for y in equation (1)

5x + 3 (8 - 4x) = 3
5x + 24 - 12x = 3
-7x = -21
x = 3


Substitute 3 for x in equation (2)

y = 8 - 4(3)
y = -4


Check:
5x + 3y = 3
5(3)+ 3(-4)= 3
3 = 3 /


y = 8 - 4x
-4 = 8 - 4(3)
-4 = -4 /

The solution is the ordered pair (3,-4).


Solution by Elimination

This method of solving systems of equations depends on the addition property of equality, which states that when two quantities are equal, adding the same number to each quantity results in equal sums.This is used when coefficient of one of the unknown quantities in the two equations are equal in absolute value.

To solve a system of linear equation in two variable:

1. Arrange the two given equations in standard form.

2. Reduce the two equations in two variables to a single equation in one variable.

a. If both equations have a variable with the same numerical coefficients, eliminate it by:

(1) Adding the two equations if there coefficients have the same opposite signs.
(2) Subtracting the two equations if there coefficients have the same sign.

b. When neither variable has the same numerical coefficients multiply one or both of the given equations by numbers which will make the numerical coefficients of one if the variables the same in both equations and proceed as in (a).

3. Solve the resulting equation for the value of the remaining variable.

4. Substitute this value in one of the given equations, to find the value of the other variable.

5. The resulting number pair is the solution.

6. Check the values in two given equations.


A. Solve the system:
x + y = 8 (1)
x - y = 2 (2)

Solution:
(x + y)+(x - y) = 8 + 2
2x = 10
x = 5

Take equation (1),
x + y = 8
5 + y = 8
y = 3

Check:
Substitute the ordered pair in both equations.
x + y = 8
5 + 3 = 8
8 = 8 /


x - y = 2
5 - 3 = 2
2 = 2 /

The solution is the ordered pair (5,3).


Solution by Comparison

This method of solving of systems of linear equations uses the transitive property of equality, which states that quantities equal to the same quantity are equal. This is used when the same unknown in the two equations has coefficient 1. The value of the unknown can then be found.

A. Solve the system:
x + 2y = 4 (1)
x - 3y = -1 (2)

Solution:

a. Solve both equations for the variable x to find y.
x = -2y + 4 (1)
x = 3y - 1 (2)

Equate equation (1) and (2) since both are equal to x.
-2y + 4 = 3y - 1
-5y = -5
y = 1

b. Solve both equations for the variable y to find x.
2y = -x 4 (3)
y = -1/2x + 2

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

Equate equations (3) and equation (4) since both are equal to y.
-1/2x +2 = 1/3x + 1/3

Multiply both sides of the equation by 6.
6 (-1/2x +2) = (1/3x + 1/3)6
-3x + 12 = 2x + 2
x = 2

Check:
x + 2y = 4
(2) + 2(1) = 4
4 = 4 /

x - 3y = -1
(2) - 3(1) = -1
-1 = -1 /

The solution is the ordered pair (2,1).


Reference:
Intermediate Algebra
Authors: Soledad Jose-Dilao Ed.D.
Julieta G. Bernabe