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Can someone pls solve this for me using substitution

Can Someone Pls Solve This For Me Using Substitution class=

Sagot :

SalmonWorldTopper

[tex] \bf \underline{★Given-} \\ [/tex]

[tex]\textsf{4x + 5y = 8};[/tex]

[tex]\textsf{5x + 4y = 12}[/tex]

[tex] \bf \underline{★To\: find-} \\ [/tex]

[tex]\textsf{the value of x and y in equation?}[/tex]

[tex] \bf \underline{★Solution-} \\ [/tex]

[tex] \sf \leadsto 4x + 5y = 8 - - - (i)[/tex]

[tex] \sf \leadsto 5x + 4y = 12 - - - (ii)[/tex]

By first equation,

[tex] \sf \leadsto 4x + 5y = 8[/tex]

[tex] \sf \leadsto 4x = 8 - 5y[/tex]

[tex] \sf \leadsto x = \dfrac{8 - 5y}{4} [/tex]

[tex]\textsf{Now, we can find the original value of Y.}\\[/tex]

[tex] \sf \leadsto 5x + 4y = 12[/tex]

[tex] \sf \leadsto 5 \bigg( \dfrac{8 - 5y}{4} \bigg) + 4y = 12[/tex]

[tex] \sf \leadsto \dfrac{40 - 25y}{4} + 4y = 12[/tex]

[tex] \sf \leadsto \dfrac{40 - 25y + 16y}{4} = 12[/tex]

[tex] \sf \leadsto \dfrac{40 - 9y}{4} = 12[/tex]

[tex] \sf \leadsto 40 - 9y = 12(4)[/tex]

[tex] \sf \leadsto 40 - 9y = 48[/tex]

[tex] \sf \leadsto -9y = 48 - 40[/tex]

[tex] \sf \leadsto -9y = 8[/tex]

[tex] \sf \leadsto y = \dfrac{ -8}{9} [/tex]

[tex]\textsf{Now, we can find the original value of X.}\\[/tex]

[tex] \sf \leadsto 4x + 5y = 8[/tex]

[tex] \sf \leadsto 4x + 5 \bigg( \dfrac{ -8}{9} \bigg) = 8[/tex]

[tex] \sf \leadsto 4x - \dfrac{40}{9} = 8[/tex]

[tex] \sf \leadsto \dfrac{36x - 40}{9} = 8[/tex]

[tex] \sf \leadsto 36x - 40 = 8(9)[/tex]

[tex] \sf \leadsto 36x - 40 = 72[/tex]

[tex] \sf \leadsto 36x = 72 + 40[/tex]

[tex] \sf \leadsto 36x = 112[/tex]

[tex] \sf \leadsto x = \dfrac{112}{36} [/tex]

[tex] \underline{\textsf{Answer-}}\\[/tex]

Therefore, the values of x and y are -8/9 and 112/36 respectively.

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