Question

Solve the system by elimination,
x - y = 11
2x + y = 19​

Answer 1

x = 14 and y = 3

Explanation:

Let

`{ \purple{ \tt{x - y = 11}}}→{ \red{ \tt{ {eq}^(n) (1)}}}`

`{ \purple{ \tt{2x + y = 19}}}→{ \red{ \tt{ {eq}^(n) (2)}}}`

Multiply Eqⁿ (1) by 2, then Eqⁿ (1) becomes

`{ \purple{ \tt{2x - 2y = 22}}}`

Subtract Eqⁿ (1) and (2)

`{ \purple{ \tt{2x - 2y = 22}}}`

`{ \purple{ \tt{2x + y \: = 19}}}`

`{ \red{ \tt{( - ) \: ( - ) \: \: \: ( - )}}}`

___________________

`{ \purple{ \tt{ \: \: \: \: \: \: \: \: \: y \: = 3}}}`

Substitute the value of y in Eqⁿ (1), then

`{ \purple{ \tt{x - 3 = 11}}}`

`{ \purple{ \tt{x = 11+3}}}`

`{ \purple{ \tt{x = 14}}}`

Answer 2

(10, - 1 )

Explanation:

x - y = 11 → (1)

2x + y = 19 → (2)

add (1) and (2) term by term to eliminate y

3x + 0 = 30

3x = 30 ( divide both sides by 3 )

x = 10

substitute x = 10 into either of the 2 equations and solve for y

substituting into (2)

2(10) + y = 19

20 + y = 19 ( subtract 20 from both sides )

y = - 1

solution is (10, - 1 )