Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)
Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)
Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer: {r = -17/26
{s = 23/13 {t = 21/26
Test a pair from each table by substituting their values into the given equation and solving.
A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1
B) y = 2x + 1 .......... 1 = 2(0) + 1 .......... 1 = 1
C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1
D) y = 2x + 1 .......... -2 = 2(0) + 1 .......... -2 ≠ 1
The only pair that satisfied the equation was from answer choice B. Therefore, B is the correct answer.
Answer:
Adding 12 to the circle area is equal to the square area.
Or
s2 = 12 + A
Where
s = side of square
A = area of circle
So
s2 = 12 + 36
s2 = 48
Solve this for s to get the side length
