Home > CCAA8 > Chapter 5 Unit 6 > Lesson CCA: 5.3.2 > Problem5-107

5-107.

Solve each equation.

1.  $(x+2)(x+3)=x^2-10$

Rewrite the left side. Then isolate $x$.

$x=-\frac{16}{5}$

1.  $\frac { 1 } { 2 } x + \frac { 1 } { 3 } x - 7 = \frac { 5 } { 6 } x$

Multiply all terms by $6$ (including the $7$), then solve.

1.  $| 2 x - 1 | = 9$

Note the absolute value sign. You will need to solve two equations:
$2x − 1 = 9$ and $2x − 1 = −9$

1.  $\frac { x + 1 } { 3 } = \frac { x } { 2 }$

$x = 2$