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solmaris [256]
3 years ago
12

Which value of x makes the equation 2/3x-3=11 true?

Mathematics
1 answer:
harkovskaia [24]3 years ago
7 0

Answer:

x = 21

Step-by-step explanation:

To see for what value of x is the equation true, we have to solve x.

\displaystyle \frac{2}{3} x - 3 = 11\\\\Adding \ 3 \ to \ both \ sides\\\\\frac{2x}{3} = 11 + 3\\\\\frac{2x}{3} = 14\\\\Multiply \ 3 \ to \ both \ sides\\\\2x = 14 \times 3\\\\2x = 42 \\\\Divide\ both\ sides\ by\ 2\\\\x = 42/2\\\\x = 21\\\\For \ x = 21\ the \ given \ equation \ is \ true.\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
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Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
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Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

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