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icang [17]
4 years ago
11

Simplify to create an equivalent expression. 6(7 – 3y)+6(y+1) Choose 1 answer:

Mathematics
1 answer:
Rus_ich [418]4 years ago
4 0

The equivalent expression is -12y + 48 or -12(y - 4) or 12(-y + 4)

<em><u>Solution:</u></em>

Given that we have to simplify to create an equivalent expression

<em><u>Given expression is:</u></em>

6(7 – 3y) + 6(y + 1)

Remove the parenthesis by multiplying terms inside the bracket with constant outside the bracket

\rightarrow 6(7 - 3y) + 6(y + 1)\\\\\rightarrow 42 - 18y + 6y + 6

Combine the like terms

\rightarrow -12y + 42 + 6

Add the constants

\rightarrow -12y + 48

Thus the equivalent expression is -12y + 48 or -12(y - 4) or 12(-y + 4)

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Answer:

x = 2 and y = 1

Step-by-step explanation:

4*2 = 8

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<em>Answer</em>

= 11

<em>Explanation</em>

f(x) = x² + 3x- 7

To get the value of f(x) when x = 3, we substitute x with 3 in the function f(x).

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Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ
const2013 [10]

Answer:

\frac{(2a + 1)^2}{50a}

Step-by-step explanation:

Given

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}

Required

Find the equivalent

We start by changing the / to *

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}

\frac{2a + 1}{10a - 5} * \frac{4a^2 - 1}{10a}

Factorize 10a - 5

\frac{2a + 1}{5(2a - 1)} * \frac{4a^2 - 1}{10a}

Expand 4a² - 1

\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1}{10a}

\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1^2}{10a}

Express (2a)² - 1² as a difference of two squares

Difference of two squares is such that: a^2- b^2= (a+b)(a-b)

The expression becomes

\frac{2a + 1}{5(2a - 1)} * \frac{(2a - 1)(2a + 1)}{10a}

Combine both fractions to form a single fraction

\frac{(2a + 1)(2a - 1)(2a + 1)}{5(2a - 1)10a}

Divide the numerator and denominator by 2a - 1

\frac{(2a + 1)((2a + 1)}{5*10a}

Simplify the numerator

\frac{(2a + 1)^2}{5*10a}

\frac{(2a + 1)^2}{50a}

Hence,

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1} = \frac{(2a + 1)^2}{50a}

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