should you perform
you multiply by the
denominator of the
fraction in the equation,
you get a one-step
equation. So, multiplying
first gets rid o f the
Why isn't —1= — 5
When you solve for a
variable, the coefficient
must be 1 , n o t - 1.
W hen o n e side of a n eq u a tio n is a fraction w ith m ore th a n one term in th e num erator,
you can still u n d o division by m ultiplying each side by th e denom inator.
Solving W ith Two Terms in the Num erator
What is the solution of x Q 7 = —12?
3( 12 )
7 = —36
7 =—3 6 + 7
^ G o t It? 3. a. W hat is th e solution of 6
M u ltip ly each side by 3.
A dd 7 to each side.
X - 2 0
b. Reasoning W rite th e right side of th e eq u a tio n in p art (a) as the
difference of two fractions. Solve th e equation. Did you find th e equation
in p art (a) or th e rew ritten eq u a tio n easier to solve? Why?
W hen you use deductive reasoning, you m ust state your steps an d your reason for each
step using properties, definitions, or rules. In P roblem 4, you are asked to provide the
reasons for each step of th e problem using deductive reasoning.
Problem 4 Using Deductive Reasoning
What is the solution of —t + 8 = 3? Justify each step.
1 +COII CO
Original equ atio n
—f + 8 — 8 = 3 — 8 S ubtraction P roperty of Equality
-f= -5 Use su b tractio n to simplify.
-11= -5 Multiplicative Property of — 1
-1 -1 Division Property of Equality
t= 5 Use division to simplify.
& G o t It? 4. W hat is th e solution of | - 5 = 4? Justify each step.
90 Ch ap t er 2 So lvin g Eq u at i o n s