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DochEvi [55]
3 years ago
6

What is the value of n in the equation 1/2(n – 4) – 3 = 3 – (2n + 3)?

Mathematics
2 answers:
Travka [436]3 years ago
8 0
Distribute 1/2 to <span>(n – 4):

</span>\frac{1}{2} \times n = \frac{1}{2}n
\frac{1}{2} \times -4 = -2
<span>
Subtract (2n + 3):

</span>- (2n + 3) = -2n -3
<span>
Your equation should now look like this:

</span>\frac{1}{2}n - 2 - 3 = 3 - 2n - 3
<span>
Combine like terms on both sides:

</span>-2 - 3 = -5
3 - 3 = 0
\frac{1}{2}n - 5 = -2n
<span>
Subtract 1/2n from both sides:

</span>-5 = -\frac{5}{2}n
<span>
Divide both sides by -5/2 to get n by itself:

</span>n = 2
<span>
The value of n is 2.</span>
Jobisdone [24]3 years ago
6 0
\dfrac{1}{2}(n-4)-3=3-(2n+3)\\\\\\\dfrac{1}{2}(n-4)-3=3-2n-3\\\\\\\dfrac{1}{2}(n-4)-3=-2n\qquad|\cdot2\\\\\\n-4-6=-4n\\\\n-10=-4n\qquad|+10\\\\n-10+10=-4n+10\\\\n=-4n+10\qquad|+4n\\\\n+4n=-4n+4n+10\\\\5n=10\qquad|:5\\\\\boxed{n=2}
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Answer:  3a = (0, 1)       3b = (2, 1)      3c = (2.5, 1)      3d = (1.6, 1)

               4a = (2, 3.5)             4b = (2, 3)           4c = (2, 5.375)

<u>Step-by-step explanation:</u>

The length of AB is 6 and is horizontal (affects the x-coordinate)

3a)\quad 6\bigg(\dfrac{1}{2}\bigg)\quad =3\qquad \qquad A(-2, 1) +(3, 0)= C(1,1)\\\\\\3b)\quad 6\bigg(\dfrac{2}{3}\bigg)\quad =4\qquad \qquad A(-2, 1) +(4,0)= C(2,1)\\\\\\3c)\quad 6\bigg(\dfrac{3}{4}\bigg)\quad =\dfrac{9}{2}\qquad \qquad A(-2, 1) +(4.5,0)= C(2.5,1)

3d)\quad 6\bigg(\dfrac{3}{5}\bigg)\quad =\dfrac{18}{5}\qquad \qquad A(-2, 1) +(3.6,0)= C(1.6,1)

***********************************************************************************

The length of AB is 5 and is vertical (affects the y-coordinate)

4a)\quad 5\bigg(\dfrac{1}{2}\bigg)\quad =\dfrac{5}{2}\qquad \qquad A(2, 1) +(0,2.5)= C(2,3.5)\\\\\\4b)\quad 5\bigg(\dfrac{2}{5}\bigg)\quad =2\qquad \qquad A(2, 1) +(0,2)= C(2,3)\\\\\\4c)\quad 5\bigg(\dfrac{7}{8}\bigg)\quad =\dfrac{35}{8}\qquad \qquad A(2, 1) +(0,4.375)= C(2,5.375)

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3 years ago
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