### Home > ACC7 > Chapter 10 Unit 11 > Lesson CC3: 10.1.1 > Problem10-17

10-17.

Solve the following equations for the given variable.

1. $3x + (−4) = 2y + 9$ for $y$

$3x + (−4) = 2y + 9$
$−9 −9$

$\frac{3x}{2}-\frac{13}{2}=\frac{2y}{2}$

$1.5x − 6.5 = y$

1. $12 =\frac { 6 } { 7 }x$ for $x$

Isolate the variable.

Multiply both sides of the equation by $\frac{7}{6}$.

$(\frac{7}{6})12=\frac{6}{7}x(\frac{7}{6})$

$x=\frac{7}{6}(12)=14$

1. $\frac { 38 } { 6 } = \frac { x } { 18 }$ for $x$

You can use cross multiplication to solve this equation.

$38 (18) = 6 (x)$

$x=\frac{684}{6}=114$

1. $\frac { 9 } { x } = \frac { 85 } { 10 y }$ for $y$

See part (c).