### Home > CCA2 > Chapter 1 > Lesson 1.2.2 > Problem1-93

1-93.

Suppose you want to find where the lines $y=3x+15$ and $y=3−3x$ cross, and you want to be more accurate than the graphing calculator or graph paper will allow. You can use algebra to find the point of intersection.

1. If you remember how to do this, find the point of intersection using algebra and be prepared to explain your method to your team tomorrow in class. If you do not remember, then do parts (b) through (e) below.

2. Since $y=3x+15$ and $y=3−3x$ , what must be true about $3x+15$ and $3−3x$ when their $y$-values are the same?

They must be the same.

3. Write an equation that does not contain $y$ and solve it for $x$.

Substitute one equation's $y$ for the other: $3x+15=3−3x$

$6x+15=3$

$6x=−12$

$x=−2$

4. Use the $x$-value you found in part (c) to find the corresponding $y$-value.

Substitute that $x$-value into one of the $y$-value equations.

$y=3(−2)+15$

$y=−6+15$

$y=9$

5. Where do the two lines cross?

$(−2,9)$