Algebraic Techniques to Solve System of Equations

Introduction to System of Equations

A system of equations is a set or collection of equations that are dealt together. These equations can be solved graphically and algebraically. The point where two lines intersect is the solution to the system of equations. This article elaborates on the algebraic technique to solve system of equations.

System of Linear Equations Solutions

A linear equation is one which has one or more variables and, when plotted, gives a straight line. Two or more linear equations consisting of two or more variables, such that all equations are considered at the same time, is called the system of linear equations.

Solution Type

1. A unique solution exists if a numerical value for each variable is found that will satisfy the system of equations.

2. Some linear equations may not have a solution or have infinitely many solutions.

3. A consistent system of equations will have at least 1 solution, whereas a system with no solution is an inconsistent system.

Methods to Solve System of Equations

The different methods to solve the system of equations are discussed below:

Substitution Method: While finding the solution to the system of linear equations having decimal or fraction values, the method of substitution is more accurate. It is the process of solving for one variable and then plugging its value of it into the second equation to obtain the value of the second variable and solve the given system. Then check the solution in both equations.

Elimination Method: It is also called the addition method. The two terms with the same variable are added to the opposite coefficients to make the sum 0. However, not all systems will have the two terms of one variable with opposite coefficients. Then cross multiplication of one or both equations is done to eliminate one of the variables.

• Arrange the variable to the left and the constants to the right.
• Write the equations according to their respective variables.
• If the variable in the first equation has the opposite coefficient in the second equation, then add them and eliminate the variable.
• If there are no opposite coefficients, cross multiply a number to the equations to obtain one to add them and eliminate one variable.
• Solve the equation for the other variable and find its value.
• Substitute the value of this variable in one of the equations and get the value of another variable.
• Then check the solution.

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Solved Examples of System of Equations

Example 1: Solve the system of equations 5x + 3y = 7, 3x – 5y = -23 by substitution method.

Solution:

5x + 3y = 7

3x – 5y = -23

Consider the first equation,

5x + 3y = 7

Subtract 3y from both sides.

5x + 3y – 3y = 7 – 3y

Divide both sides by 5.

On simplification,

Substitute

For x = (7 – 3y)/5

Substitute y = 4,

The solutions to the system of equations are y = 4 and x = -1.

Example 2: Solve the system of equations x + 2y = −1; −x + y = 3 by elimination method.

Solution:

Solve 3y = 2 for y.

3y = 2

y = 2/3

For x + 2y = -1

Plugin y = 2/3 x + 2 . 2/3 = −1

Subtract 8/3 from both sides, and we get

x + 2 ⋅ 2/3 − 2 ⋅ 2/3 = −1 − 2 ⋅ 2/3

or x = −7/3

The solutions to the system of equations are: y = 2/3 and x = -7/3

Example 3: Solve the pair of equations 8x + 5y = 23; 3x + 2y = 9 by elimination method.

Solution:

Given equations are 8x + 5y = 23 ..(1)

3x + 2y = 9 ….(2)

Multiply (1) by 3.

24x + 15y = 69 ..(3)

Multiply (2) by 8.

24x + 16y = 72 ….(4)

Subtract (4) from (3).

24x + 15y = 69

-24x – 16y = -72

⇒ -y = -3

⇒ y = 3

Substitute the value of y in (2).

3x + 2 × 3 = 9

3x + 6 = 9

3x = 9 – 6

3x = 3

⇒ x = 3/3 = 1

Hence, x = 1 and y = 3.

Example 4: What are the number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3) y =3k − 1 has infinitely many solutions?

Solution:

For infinitely many solutions, the two equations must be identical.

⇒ [k + 1] / [k] = 8 / [k + 3] = 4k / [3k − 1]

⇒ (k + 1) (k + 3) = 8k and 8 (3k − 1) = 4k (k + 3)

⇒ k²− 4k + 3 = 0 and k² − 3k + 2 = 0.

By cross multiplication, k² / [− 8 + 9] = k / [3−2] = 1 / [−3+4] k² = 1 and k = 1.

Example 5: Solve by substitution method.

Solution:

Isolate x for x + 3y + 3z = 53.

x + 3y + 3z = 53

Subtract 3y + 3z from both sides.

x + 3y + 3z – (3y + 3z) = 53 – (3y + 3z)

x = 53 – 3y – 3z

Substitute x = 53 – 3y – 3z

Isolate y for 53 – 3y – 3z + y – 3z = 4.

53 – 3y – 3z + y – 3z = 4

-2y + 53 = 4 + 6z

-2y + 53 – 53 = 4 + 6z – 53

-2y = 6z – 49

y = -(6z – 49)/2

Substitute y = -(6z – 49)/2.

Example 6: Solve by elimination method. 

-y – 2z = -8

x + 3z = 2

7x + y + z = 0

Solution:

-y – 2z = -8

x + 3z = 2

7x + y + z = 0

Arrange equation variables for elimination.

Rearrange equations.

[y + z + 7x = 0] + [-y – 2z = -8] = [-z + 7x = -8]

Multiply -z + 7x = -8 by 3.

[3z + x = 2] + [-3z + 21x = -24] = [22x = -22]

22x = -22

x = -1

-3z + 21(-1) = -24

-3z = -3

z = 1

-y – (2)(1) = -8

y = 6

The solutions to the system of equations are: x = -1, z = 1, y = 6