Page 679 - Algebra 1

Example
3
664 Saxon Algebra 1
x-1 x 2x _3 - _2 _5
3-1 3 2(3) _5 = _5 ✓
66
The solution is x = 3.
Substitute 3 for x in the original equation. Simplify.
Using the LCD to Solve Subtraction Equations
Solve _3 - _2 = _5 . x-1 x 2x
SOLUTION
_3 -_2=_5 x-1 x 2x
Caution
Remember to distribute the negative over the new numerator in a fraction following a subtraction sign.
The LCD is 2x(x - 1).
2x(x - 1)_3 -2x(x - 1)_2
= 2x(x - 1)_5 2x
=5(x-1) =5x-5
=5x-5 =3x-5
=3x =x
Multiply each term by the LCD.
Simplify each term.
Use the Distributive Property.
Collect like terms.
Subtract 2x from both sides.
Add 5 to both sides. Divide both sides by 3.
x-1 x 6x-4(x-1)
6x - 4x + 4 2x + 4
4
9 3
Check Verify that the solution is not extraneous. _3 -_2=_5
Example
4
Solve the equation.
x-1= x+9 __
x - 2 2x - 4 SOLUTION
Checking for Extraneous Solutions
Use cross products. Multiply.
Subtract x2, 7x, and -18 from both sides.
Factor.
Use the Zero Product Property to solve.
x-1= x+9 __
x - 2 2x - 4
(x - 1)(2x - 4) = (x - 2)(x + 9)
2x2 -6x+4=x2 +7x-18 x2 -13x+22=0
(x - 11)(x - 2) = 0 x = 11 or x = 2