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# Graphical Solution to System of Equations Help

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## Introduction to Graphical Solution to System of Equations

Two lines in the plane either intersect in one point, are parallel, or are really the same line. Until now, our lines have intersected in one point. When solving a system of two linear equations that are parallel or are on the same line, both variables will cancel and we are left with a true statement such as “3 = 3” or a false statement such as “5 = 1.” We will get a true statement when the two lines are the same and a false statement when they are parallel.

#### Examples

Fig. 10.4

This is a false statement, so the lines are parallel. They are sketched in Figure 10.4

• We will use substitution.

Because 0 = 0 is a true statement, these lines are the same.

When the system of equations is not a pair of lines, there could be no solutions, one solution, or more than one solution. The same methods used for pairs of lines will work with other kinds of systems.

### Examples

• Elimination by addition would not work to eliminate x 2 because B has no x 2 term to cancel x 2 in A. Solving for x in B and substituting it in for x in A would work to eliminate x . Both addition and substitution will work to eliminate y . We will use addition to eliminate y .

Fig. 10.5

The solutions occur when x = 1 or x = 4. We need to find two y -values. We will let x = 1 and x = 4 in A.

y = 1 2 − 2(1) − 3 = −4 (1, −4) is one solution.

y = 4 2 − 2(4) − 3 = 5 (4, 5) is the other solution.

We can see from the graphs in Figure 10.5 that these solutions are correct.

• We could solve for x 2 in A and substitute this in B. We cannot add the equations to eliminate y or y 2 because A does not have a y term to cancel y in B and B does not have a y 2 term to cancel y 2 in A. We will move to the left side of B and multiply B by −3. Then we can add this to A to eliminate x 2 .

The solutions occur when y = 4, −1. Put y = 4, −1 in A to find their x -values.

x 2 + 4 2 = 25

x 2 = 9

x = ± 3 (−3, 4) and (3,4) are solutions.

x 2 + (− 1) 2 = 25

x 2 = 24

• Addition will not work on this system but substitution will. We will substitute for y in A.

## Systems of Equations and Inequalities Practice Problems

#### Practice

Solve the systems of equations. Put your solutions in the form of a point, ( x , y ).

1.

2.

3.

4.

#### Solutions

1.

There are solutions for x = −4 and x = 3. Put these in A.

y = (−4) 2 − 4 = 12; (−4, 12) is a solution.

y = 3 2 − 4 = 5; (3, 5) is a solution.

2. Substitute −2 x − 5 for y in A.

x 2 + (−2 x − 5) 2 + 6 x − 2(−2 x − 5) = −5

x 2 + 4 x 2 + 20 x + 25 + 6 x + 4 x + 10 = −5

5 x 2 + 30 x + 40 = 0 Divide by 5

x 2 + 6 x + 8 = 0

( x + 4)( x + 2) = 0

There are solutions for x = −4 and x = −2. We will put these in B instead of A because there is less computation to do in B.

y = −2(−4) −5 = 3; (−4, 3) is a solution.

y = −2(−2) −5 = −1; (−2, −1) is a solution.

3.

Put x = 4 and x = −4 in A.

 (−4) 2 − y 2 = 16 4 2 − y 2 = 16 16 − y 2 = 16 16−y 2 = 16 y 2 = 0 y 2 = 0 y = 0 y = 0

The solutions are (−4, 0) and (4, 0).

4. Substitute for y in A.

The solutions are (from 2 x − 1 = 0 and 2 x + 1 = 0) and x = ±1. Put these in B.

Practice problems for this concept can be found at: Systems of Equations and Inequalities Practice Test.

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