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

Solve for (x,y)

Sagot :

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[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:x = 5/2 [/tex]

[tex] \qquad \tt \rightarrow \: y = 3/2[/tex]

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[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: x = \cfrac{y}{3} + 2 \: \: \: \: \: \: \: \: - (1)[/tex]

[tex]\qquad \tt \rightarrow \: 7x -5 y = 10 \: \: - (2)[/tex]

[tex] \textsf{put value of x in equation 2} [/tex]

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

[tex]\qquad \tt \rightarrow \: \cfrac{7y}{3} + 14 - 5y = 10 [/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{7y}{3} - 5y = 10 - 14[/tex]

[tex]\qquad \tt \rightarrow \: \cfrac{7y - 15y}{3} = - 4[/tex]

[tex]\qquad \tt \rightarrow \: - 8y = - 4(3)[/tex]

[tex]\qquad \tt \rightarrow \: y = \cfrac{ - 12}{ - 8} [/tex]

[tex]\qquad \tt \rightarrow \: y = \cfrac{ 3}{ 2} [/tex]

[tex] \textsf{put value of y in equation 1 } [/tex]

[tex]\qquad \tt \rightarrow \: x = \cfrac{y}{3} + 2[/tex]

[tex]\qquad \tt \rightarrow \: x = \bigg (\cfrac{3}{2} \sdot\cfrac{1}{3} \bigg) + 2[/tex]

[tex]\qquad \tt \rightarrow \: x = \cfrac{1}{2} + 2[/tex]

[tex]\qquad \tt \rightarrow \: x = \cfrac{1 + 4}{2} [/tex]

[tex]\qquad \tt \rightarrow \: x = \cfrac{5}{2} [/tex]

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