### Home > INT3 > Chapter 1 > Lesson 1.2.1 > Problem1-72

1-72.

Rewrite each equation by solving for $x$. (Note that $y$ will be in your answer).

1. $y = \frac {3}{5}x + 1$

To solve for $x$, you need to isolate the $x$-term. There are two ways to start. You can isolate the $x$-term or eliminate the denominator.

To isolate the $x$-term:

$y-1=\frac{3}{5}x$

$\frac{5}{3}\cdot \left(y-1\right)=\frac{3}{5}x\cdot \frac{5}{3}$

$x=\frac{5(y-1)}{3}$

To eliminate the denominator first:

$5\cdot y=\left(\frac{3}{5}x+1\right)\cdot 5$

$5y=3x+5$

$3x=5y-5$

$x=\frac{5y-5}{3}$

1. $3x + 2y = 6$

See part (a).

1. $y = x^2$

Refer to part (c) of problem 1-37.

1. $y = x^2 - 100$

Refer to part (d) of problem 1-37.

This cannot be simplified. Why?

$x=\pm\sqrt{y+100}$