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# Solving Systems of Equations and Inequalities Help

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Updated on Oct 27, 2011

## Solving a Sytstem of Equations Using Substitution and Linear Combination

A system of equations is a set of two or more equations with the same solution. For example, if we're told that both 2c + d = 11 and c + 2d = 13 have the same solution, that means the variables c and d have the same value in both equations.

Two methods for solving a system of equations are substitution and linear combination.

### Substitution

Substitution involves solving for one variable in terms of another and then substituting that expression into the second equation.

#### Example

Here are those two equations with the same solution:

2c + d = 11 and c + 2d = 13

To solve, first choose one of the equations and rewrite it, isolating one variable in terms of the other. It does not matter which variable you choose.

2c + d = 11 becomes d = 11 – 2c

Next substitute 11 – 2c for d in the other equation and solve:

c + 2d = 13

c + 2(11 – 2c) = 13

c + 22 – 4c = 13

22 – 3c = 13

22 = 13 + 3c

9 = 3c

c = 3

Now substitute this answer into either original equation for c to find d.

2c + d = 11

2(3) + d = 11

6 + d = 11

d = 5

Thus, c = 3 and d = 5.

### Linear Combination

Linear combination, which can also be used to solve a system of equations, involves writing one equation over another and then adding or subtracting the like terms so that one letter is eliminated.

#### Example

x – 7 = 3y and x + 5 = 6y

First rewrite each equation in the same form.

x – 7 = 3y becomes x – 3y = 7

x + 5 = 6y becomes x – 6y = –5.

Now subtract the two equations so that the x terms are eliminated, leaving only one variable:

y = 4 is the answer.

Now substitute 4 for y in one of the original equations and solve for x.

x – 7 = 3y

x – 7 = 3(4)

x – 7 = 12

x – 7 + 7 = 12 + 7

x = 19

So, the solution to the system of equations is y = 4 and x = 19.

### Systems of Equations With No Solution

It is possible for a system of equations to have no solution if there are no values for the variables that would make all the equations true. For example, the following system of equations has no solution because there are no values of x and y that would make both equations true:

3x + 6y = 14

3x + 6y = 9

In other words, the same expression cannot equal both 14 and 9.

## Graphing Systems of Equalities

To graph a system of linear equations, use what you already know about graphing linear equations. To graph a system of linear equations, you will use the slope-intercept form of graphing. The first step is to transform the equations into slope-intercept form or y = mx + b. Then use the slope and y-intercept to graph the line. Once you have both lines graphed, determine your solutions.

#### Example

 x – y = 6 2x + y = 3 Transform the first equation into y = mx + b. x – y = 6 Subtract x from both sides of the equation. x – x – y = 6 – x Simplify. –y = 6 – x Use the commutative property. –y = –x + 6 Multiply both sides of the equation by –1. –1 · –y = –1(–x + 6) Simplify both sides. y = x – 6 Transform the second equation into y = mx + b. 2x + y = 3 Subtract 2x from both sides of the equation. 2x – 2x + y = 3 – 2x Simplify. y = 3 – 2x Use the commutative property. y = –2x + 3

The slope of the first equation is 1, and the y-intercept is –6. The slope of the second equation is –2, and the y-intercept is 3. In the first equation, the line cuts the y-axis at –6. From that point, go up 1 and to the right 1. Draw a line through your beginning point and the endpoint—this line can extend as long as you want in both directions because it is endless. In the second equation, the line cuts the y-axis at 3. From that point, go down 2 and to the right 1. Draw a line through your beginning point and the endpoint, extending it as long as you want. The point of intersection of the two lines is (3,–3), so there is one solution.

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