Answer:
x = 3, y = -4
Step-by-step explanation by substitution:
Solve the following system:
{4 x + 3 y = 0 | (equation 1)
5 y + 53 = 11 x | (equation 2)
Express the system in standard form:
{4 x + 3 y = 0 | (equation 1)
-(11 x) + 5 y = -53 | (equation 2)
Swap equation 1 with equation 2:
{-(11 x) + 5 y = -53 | (equation 1)
4 x + 3 y = 0 | (equation 2)
Add 4/11 × (equation 1) to equation 2:
{-(11 x) + 5 y = -53 | (equation 1)
0 x+(53 y)/11 = -212/11 | (equation 2)
Multiply equation 2 by 11/53:
{-(11 x) + 5 y = -53 | (equation 1)
0 x+y = -4 | (equation 2)
Subtract 5 × (equation 2) from equation 1:
{-(11 x)+0 y = -33 | (equation 1)
0 x+y = -4 | (equation 2)
Divide equation 1 by -11:
{x+0 y = 3 | (equation 1)
0 x+y = -4 | (equation 2)
Collect results:
Answer: {x = 3 , y = -4
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Solve the following system:
{4 x + 3 y = 0
5 y + 53 = 11 x
Hint: | Choose an equation and a variable to solve for.
In the first equation, look to solve for x:
{4 x + 3 y = 0
5 y + 53 = 11 x
Hint: | Isolate terms with x to the left hand side.
Subtract 3 y from both sides:
{4 x = -3 y
5 y + 53 = 11 x
Hint: | Solve for x.
Divide both sides by 4:
{x = -(3 y)/4
5 y + 53 = 11 x
Hint: | Perform a substitution.
Substitute x = -(3 y)/4 into the second equation:
{x = -(3 y)/4
5 y + 53 = -(33 y)/4
Hint: | Choose an equation and a variable to solve for.
In the second equation, look to solve for y:
{x = -(3 y)/4
5 y + 53 = -(33 y)/4
Hint: | Isolate y to the left hand side.
Subtract 53 - (33 y)/4 from both sides:
{x = -(3 y)/4
(53 y)/4 = -53
Hint: | Solve for y.
Multiply both sides by 4/53:
{x = -(3 y)/4
y = -4
Hint: | Perform a back substitution.
Substitute y = -4 into the first equation:
Answer: {x = 3
, y = -4
. We have a value for YZ of 3; side a is XZ. That means in order to solve this we need WZ, which we can find using pythagorean's theorem. 3^2 + 4^2 = c^2 and 9 + 16 = c^2 and c = 5. Now we fill in accordingly: 
. Cross-multiply to get 3XZ=25 and side XZ is 
. XZ is 25/3 and YZ is 3, so 25/3 - 3 = XY. That means that XY (side a) = 16/3 or 5 1/3, choice B, or the second one down.