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# substitution method

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2x+4y=−32;−3x+y=6 /using substitution method

Guest Jul 12, 2017
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#1
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Solve the following system:
{2 x + 4 y = -32
y - 3 x = 6

In the second equation, look to solve for y:
{2 x + 4 y = -32
y - 3 x = 6

Add 3 x to both sides:
{2 x + 4 y = -32
y = 3 x + 6

Substitute y = 3 x + 6 into the first equation:
{2 x + 4 (3 x + 6) = -32
y = 3 x + 6

2 x + 4 (3 x + 6) = (12 x + 24) + 2 x = 14 x + 24:
{14 x + 24 = -32
y = 3 x + 6

In the first equation, look to solve for x:
{14 x + 24 = -32
y = 3 x + 6

Subtract 24 from both sides:
{14 x = -56
y = 3 x + 6

Divide both sides by 14:
{x = -4
y = 3 x + 6

Substitute x = -4 into the second equation:
Answer: | x = -4         and              y = -6

Guest Jul 12, 2017
#2
+1214
+2

To solve for and y, first solve for a variable and plug it into the other equation. Since you specified solving by substitution, I'll use that method. Here are your 2 equations:

1. \(2x+4y=-32\)

2. \(-3x+y=6\)

I'll solve for in the second equation because it has a coefficient of 1.

 \(-3x+y=6\) To isolate y, add 3x to both sides. \(y=6+3x\)

Now that I have solved for in one equation, substitute y in the other.

 \(2x+4y=-32\) In the previous calculation, we deduced that y=6+3x, so replace y. \(2x+4(6+3x)=-32\) Distribute the 4 into the parentheses to ease the simplification process. \(2x+24+12x=-32\) Combine the like terms, specifically 2x and 12x. \(14x+24=-32\) Subtract 24 on both sides. \(14x=-56\) Divide by 14 on both sides to isolate x. \(x=-4\)

Plug into an equation and solve for y. I'll plug it into equation 2

 \(-3x+y=6\) Substitute the calculated value for x, -4. \(-3(-4)+y=6\) Simplify -3*-4. \(12+y=6\) Subtract 12 on both sides \(y=-6\)

Therefore, the solution set is (-4,-6)

TheXSquaredFactor  Jul 12, 2017
#3
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x+y=97; x-y=39 / using elimination method/ SHOW WORK

Guest Jul 12, 2017
edited by Guest  Jul 12, 2017
#4
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Solve the following system:
{x + y = 97 | (equation 1)
x - y = 39 | (equation 2)
Subtract equation 1 from equation 2:
{x + y = 97 | (equation 1)
0 x - 2 y = -58 | (equation 2)
Divide equation 2 by -2:
{x + y = 97 | (equation 1)
0 x+y = 29 | (equation 2)
Subtract equation 2 from equation 1:
{x+0 y = 68 | (equation 1)
0 x+y = 29 | (equation 2)
Collect results:
Answer: | x = 68       and         y = 29