Exercise 27 Page 115

Another way to solve this type of equation — a proportion which contains algebraic expressions — is by using the Properties of Equality without starting with the Cross Products Property. Let's see how this would have worked with the given equation.

\frac{x - 3}{5} = \frac{6}{1 0}

▼

Solve for

x

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

x - 3 = \frac{6}{1 0} \cdot 5

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

x - 3 = \frac{3 0}{1 0}

CalcQuot

Calculate quotient

x - 3 = 3

AddEqn

$LHS + 3 = RHS + 3$

x = 6

This process was much faster than our original solving method! So, when is it easier to use cross multiplication and when is it faster to solve without it? The general rule of thumb is to look at where the variable is located. If the variable is in the denominator, solving is quicker with cross multiplication. If the variable is in the numerator, solving is quicker without it.

Use Cross Multiplication ? \frac{2}{x - 3} = \frac{7}{4} Probably! \frac{8 x + 1}{3} = \frac{1 1}{1 0} Maybe not ?