Question

Solve using substitution method:
X= y/3 +2
7x − 5y = 10

Solve for (x,y)

Answer 1

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

`\qquad \tt \rightarrow \:x = 5/2`

`\qquad \tt \rightarrow \: y = 3/2`

____________________________________

`\large \tt Solution \: :`

`\qquad \tt \rightarrow \: x = \cfrac{y}{3} + 2 \: \: \: \: \: \: \: \: - (1)`

`\qquad \tt \rightarrow \: 7x -5 y = 10 \: \: - (2)`

`\textsf{put value of x in equation 2}`

`\qquad \tt \rightarrow \: 7 \bigg( \cfrac{y}{3} + 2 \bigg) - 5y = 10`

`\qquad \tt \rightarrow \: \cfrac{7y}{3} + 14 - 5y = 10`

`\qquad \tt \rightarrow \: \cfrac{7y}{3} - 5y = 10 - 14`

`\qquad \tt \rightarrow \: \cfrac{7y - 15y}{3} = - 4`

`\qquad \tt \rightarrow \: - 8y = - 4(3)`

`\qquad \tt \rightarrow \: y = \cfrac{ - 12}{ - 8}`

`\qquad \tt \rightarrow \: y = \cfrac{ 3}{ 2}`

`\textsf{put value of y in equation 1 }`

`\qquad \tt \rightarrow \: x = \cfrac{y}{3} + 2`

`\qquad \tt \rightarrow \: x = \bigg (\cfrac{3}{2} \sdot\cfrac{1}{3} \bigg) + 2`

`\qquad \tt \rightarrow \: x = \cfrac{1}{2} + 2`

`\qquad \tt \rightarrow \: x = \cfrac{1 + 4}{2}`

`\qquad \tt \rightarrow \: x = \cfrac{5}{2}`