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kenny6666 [7]
2 years ago
15

How to solve -4{1-2x}-1=7x

Mathematics
2 answers:
mina [271]2 years ago
7 0

Answer:

<h2>x = 5</h2>

Step-by-step explanation:

-4(1-2x)-1=7x\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\(-4)(1)+(-4)(-2x)-1=7x\\\\-4+8x-1=7x\qquad\text{combine like terms}\\\\8x+(-4-1)=7x\\\\8x-5=7x\qquad\text{add 5 to both sides}\\\\8x-5+5=7x+5\\\\8x=7x+5\qquad\text{subtract}\ 7x\ \text{from both sides}\\\\8x-7x=7x-7x+5\\\\x=5

RoseWind [281]2 years ago
4 0

Answer:

5

Step-by-step explanation:

 ((0 -  4 • (1 - 2x)) -  1) -  7x  = 0

Step  2  :

Equation at the end of step  2  :

 x - 5  = 0

Step  3  :

Solving a Single Variable Equation :

3.1      Solve  :    x-5 = 0

Add  5  to both sides of the equation :

                     x = 5

One solution was found :

                  x = 5

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Answer:

we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

Step-by-step explanation:

Given the expression

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}

=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}

\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}

=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}

cancel the common factor: 2

=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}

cancel the common factor: p

=\frac{2p+1}{p^2\left(4p^2-1\right)}

=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}

cancel the common factor: 2p+1

=\frac{1}{p^2\left(2p-1\right)}

Expanding

=\frac{1}{2p^3-p^2}

Thus, we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

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