# Solving equations

• Aug 4th 2006, 05:07 AM
pashah
Solving equations
I would appreciate any assistance with the following problems.
• Aug 4th 2006, 05:40 AM
Soroban
Hello, pashah!

Quote:

Solve each equation and check for extraneous solutions.

$a)\;\sqrt{w - 3}\:=\:\sqrt{4w + 15}$

Square: . $w - 3\:=\:4w + 15$

Then: . $-3w\:=\:18\quad\Rightarrow\quad \boxed{w = -6}$

But when we check, we get: . $\sqrt{-6 - 3}\:=\:\sqrt{-9}$ . . . an "imaginary" number.

Therefore, this equation has no solutions.

Quote:

$b)\;\sqrt{x^2-6x=2} \:=\:\frac{x}{2}$

Square: . $x^2-5x + 2\:=\:\frac{x^2}{4}$

Multiply by 4: . $4x^2 - 20x + 8\:=\:x^2\quad\Rightarrow\quad 3x^2 - 20x + 8 \:= \:0$

Quote:

$\frac{5}{y-3} \;= \;1 + \frac{y+7}{2y-6}$

We have: . $\frac{5}{y-3} \;= \;1 + \frac{y+7}{2(y-3)}$

Multiply by the LCD: $2(y-3)$

. . $2(y-3)\cdot\frac{5}{y-3} \;\;= \;\;2(y-3)\cdot1 \;+ \;2(y-3)\cdot\frac{y+7}{2(y-3)}$

We have: . $2\cdot5 \;= \;2(y-3) + (y + 7)\quad\Rightarrow\quad 10 \;= \;2y - 6 + y + 7$

Then: . $9\:=\:3y\quad\Rightarrow\quad \boxed{y = 3}$

But when we check, it starts with: . $\frac{5}{3-3} \:= \:\frac{5}{0}$ . . . undefined!

Therefore, this equation has no solutions.