Question

Solve

5(2x-y) = 7x + 1
3(3x+y) = 5(x-y+12)
using the elimination method
thanks!

Answer 1

{x = 7 , y = 4

Step-by-step explanation using elimination:

Solve the following system:

{5 (2 x - y) = 7 x + 1 | (equation 1)

{3 (3 x + y) = 5 (12 + x - y) | (equation 2)

Express the system in standard form:

{3 x - 5 y = 1 | (equation 1)

{4 x + 8 y = 60 | (equation 2)

Swap equation 1 with equation 2:

{4 x + 8 y = 60 | (equation 1)

{3 x - 5 y = 1 | (equation 2)

Subtract 3/4 × (equation 1) from equation 2:

{4 x + 8 y = 60 | (equation 1)

{0 x - 11 y = -44 | (equation 2)

Divide equation 1 by 4:

{x + 2 y = 15 | (equation 1)

{0 x - 11 y = -44 | (equation 2)

Divide equation 2 by -11:

{x + 2 y = 15 | (equation 1)

{0 x+y = 4 | (equation 2)

Subtract 2 × (equation 2) from equation 1:

{x+0 y = 7 | (equation 1)

{0 x+y = 4 | (equation 2)

Collect results:

Answer: {x = 7 , y = 4

Answer 2

Answer: x = 7, y = 4

Explanation:

Equation 1:

5(2x - y) = 7x + 1

10x - 5y = 7x + 1 distributed 5 on the left side

3x - 5y = 1 subtracted 7x from both sides

Equation 2:

3(3x + y) = 5(x - y + 12)

9x + 3y = 5x - 5y + 60 distributed 3 on the left and 5 on the right

4x + 3y = -5y + 60 subtracted 5x from both sides

4x + 8y = 60 added 5y to both sides

Choose which variable to eliminate (I am going to eliminate x), then multiply each equation by the LCM of the coefficients so their sum is zero.

3x - 5y = 1 → -4(3x - 5y = 1) → -12x + 20y = -4

4x + 8y = 60 → 3(4x + 8y = 60) → 12x + 24y = 180

44y = 176

y = 4

Substitute y = 4 into either of the equations to solve for x:

3x - 5y = 1

3x - 5(4) = 1

3x - 20 = 1

3x = 21

x = 7

Check (substitute x = 7, y = 4 into the other equation):

4x + 8y = 60

4(7) + 8(4) = 60

28 + 32 = 60

60 = 60
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