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makkiz [27]
3 years ago
9

What kind of solution is -4x-4=-7(x+4) ?

Mathematics
2 answers:
natima [27]3 years ago
8 0

~ Simplifying

-4x + -4 = -7(x + 4)

~ Reorder the terms:

-4 + -4x = -7(x + 4)

~ Reorder the terms:

-4 + -4x = -7(4 + x)

-4 + -4x = (4 * -7 + x * -7)

-4 + -4x = (-28 + -7x)

~ Solving

-4 + -4x = -28 + -7x

~ Solving for variable 'x'.

~ Move all terms containing x to the left, all other terms to the right.

~ Add '7x' to each side of the equation.

-4 + -4x + 7x = -28 + -7x + 7x

~ Combine like terms: -4x + 7x = 3x

-4 + 3x = -28 + -7x + 7x

~ Combine like terms: -7x + 7x = 0

-4 + 3x = -28 + 0

-4 + 3x = -28

~ Add '4' to each side of the equation.

-4 + 4 + 3x = -28 + 4

~ Combine like terms: -4 + 4 = 0

0 + 3x = -28 + 4

3x = -28 + 4

~ Combine like terms: -28 + 4 = -24

3x = -24

~ Divide each side by '3'.

x = -8

~ Simplifying

x = -8

Schach [20]3 years ago
7 0
One solution (x=-8)

Work is posted below

Hope this helps comment below for more questions :)

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Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)


Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)


Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

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Step 3: Rewrite in function notation

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