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Core Connections Integrated I, 2013 View details

1. Section 4.1

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Exercise 29 Page 206

a To solve the equation we must eliminate the fractions. Since the coefficient of the fraction on the left-hand side has a denominator of

7,

we should multiply both sides by

7 .

We can also multiply by

10

to eliminate the second fraction.

\frac{2}{7} x + 2 = \frac{3}{1 0}

MultEqn

$LHS \cdot 7 = RHS \cdot 7$

2 x + 14 = \frac{2 1}{1 0}

MultEqn

$LHS \cdot 10 = RHS \cdot 10$

20 x + 140 = 21

The equation

20 x + 140 = 21

is much easier to solve.

20 x + 140 = 21

SubEqn

$LHS - 140 = RHS - 140$

20 x = - 119

DivEqn

$LHS / 20 = RHS / 20$

x = - \frac{1 1 9}{2 0}

To check our solution, we should substitute it into the original equation and simplify. If the left-hand side and right-hand side are equal, we have solved the equation correctly.

\frac{2}{7} x + 2 = \frac{3}{1 0}

Substitute

$x = - \frac{1 1 9}{2 0}$

\frac{2}{7} (- \frac{1 1 9}{2 0}) + 2 = \frac{3}{1 0}

▼

Simplify left-hand side

MultPosNeg

$a (- b) = - a \cdot b$

- \frac{2}{7} (\frac{1 1 9}{2 0}) + 2 = ? \frac{3}{1 0}

MultFrac

Multiply fractions

- \frac{2 \cdot 1 1 9}{7 \cdot 2 0} + 2 = ? \frac{3}{1 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 4}{b / 1 4}$

- \frac{1 7}{1 0} + 2 = ? \frac{3}{1 0}

CommutativePropAdd

Commutative Property of Addition

2 - \frac{1 7}{1 0} = ? \frac{3}{1 0}

NumberToFrac

$a = \frac{1 0 \cdot a}{1 0}$

\frac{2 0}{1 0} - \frac{1 7}{1 0} = ? \frac{3}{1 0}

SubFrac

Subtract fractions

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

As we can see, the solution is correct.

b Notice that all of the terms are fractions. Examining them, we notice that they are all multiples of

50 .

Therefore, we can solve for

x

by multiplying both sides by

50 .

\frac{9}{5 0} + \frac{6 x}{2 5} = \frac{3}{1 0}

MultEqn

$LHS \cdot 50 = RHS \cdot 50$

\frac{9}{5 0} \cdot 50 + \frac{6 x}{2 5} \cdot 50 = \frac{3}{1 0} \cdot 50

MoveRightFacToNum

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

\frac{9 \cdot 5 0}{5 0} + \frac{6 x \cdot 5 0}{2 5} = \frac{3 \cdot 5 0}{1 0}

CalcQuot

Calculate quotient

9 + 6 x \cdot 2 = 3 \cdot 5

Multiply

9 + 12 x = 15

The equation

9 + 12 x = 15

is much easier to solve.

9 + 12 x = 15

SubEqn

$LHS - 9 = RHS - 9$

12 x = 6

DivEqn

$LHS / 12 = RHS / 12$

x = \frac{6}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 6}{b / 6}$

x = \frac{1}{2}

To check our solution, we should substitute it into the original equation and simplify. If the left-hand side and right-hand side are equal, we have solved the equation correctly.