Solve the system of linear equations. separate the x- and y- with a coma.
2 answers:
Solve the following system:
{6 x = -8 y - 14 | (equation 1)
{-12 x = 8 y + 20 | (equation 2)
Express the system in standard form:
{6 x + 8 y = -14 | (equation 1)
{-(12 x) - 8 y = 20 | (equation 2)
Swap equation 1 with equation 2:
{-(12 x) - 8 y = 20 | (equation 1)
{6 x + 8 y = -14 | (equation 2)
Add 1/2 × (equation 1) to equation 2:
{-(12 x) - 8 y = 20 | (equation 1)
{0 x+4 y = -4 | (equation 2)
Divide equation 1 by 4:
{-(3 x) - 2 y = 5 | (equation 1)
{0 x+4 y = -4 | (equation 2)
Divide equation 2 by 4:
{-(3 x) - 2 y = 5 | (equation 1)
{0 x+y = -1 | (equation 2)
Add 2 × (equation 2) to equation 1:
{-(3 x)+0 y = 3 | (equation 1)
{0 x+y = -1 | (equation 2)
Divide equation 1 by -3:
{x+0 y = -1 | (equation 1)
{0 x+y = -1 | (equation 2)
Collect results:
Answer: {x = -1
{y = -1
Please note the parentheses should span over both equations but the editor doesn't allow that. see example.
{6 x = -8 y - 14 | (equation 1)
{-12 x = 8 y + 20 | (equation 2)
Express the system in standard form:
{6 x + 8 y = -14 | (equation 1)
{-(12 x) - 8 y = 20 | (equation 2)
Swap equation 1 with equation 2:
{-(12 x) - 8 y = 20 | (equation 1)
{6 x + 8 y = -14 | (equation 2)
Add 1/2 × (equation 1) to equation 2:
{-(12 x) - 8 y = 20 | (equation 1)
{0 x+4 y = -4 | (equation 2)
Divide equation 1 by 4:
{-(3 x) - 2 y = 5 | (equation 1)
{0 x+4 y = -4 | (equation 2)
Divide equation 2 by 4:
{-(3 x) - 2 y = 5 | (equation 1)
{0 x+y = -1 | (equation 2)
Add 2 × (equation 2) to equation 1:
{-(3 x)+0 y = 3 | (equation 1)
{0 x+y = -1 | (equation 2)
Divide equation 1 by -3:
{x+0 y = -1 | (equation 1)
{0 x+y = -1 | (equation 2)
Collect results:
Answer: {x = -1
{y = -1
Please note the parentheses should span over both equations but the editor doesn't allow that. see example.
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