Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
1. Solving Systems by Graphing
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Exercise 33 Page 368

What does it mean that (3,7) is a solution to a system of equations?

See solution.

Practice makes perfect

To help us answer this question we will take a closer look at two different systems of equations. System I & System II y=- x+10 y=2x+1 & y=x+3.98 y=- 2x+13.01 Let's solve both systems graphically.

We have found that both systems appear to have the point ( 3, 7) as a point of intersection. However, we cannot be sure that the point really is a solution. Therefore, it is a good idea to check our work.

Checking Our Work

To check our work and be sure that we have found the correct solution to the system, we can substitute x= 3 and y= 7 into both equations. If we then end up with true statements we can be sure that the ordered pair is the solution. Let's do that for our systems. First we will check ( 3, 7) in System I.

y=- x+10 & (I) y=2x+1 & (II)
7? =- 3+10 7? =2( 3)+1
7? =- 3+10 7? =6+1

(I), (II): Add terms

7=7 7=7

We ended up with two true statements. Therefore, (3,7) is a solution to System I. Let's also check the same point in Equation II.

y=x+3.98 y=- 2x+13.01
7? = 3+3.98 7? =- 2( 3)+13.01
7? =3+3.98 7? =-6+13.01

(I), (II): Add terms

7≠ 6.98 7≠ 7.01

Here we ended up with two false statements. This means that the point we tried was not a solution.

Can We Be Sure?

If we have two true statements we can be sure that the ordered pair is the solution. If one or both is a false statement (where the left-hand side and the right-hand side of the expression are not equal) we can be sure that the ordered pair is not the solution.