Question

solve for x and y by elimination method 3x+3y=4. and 2x -3y=7​

Answer 1

Hallo dear ,

Your answer is attached below.

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Solve for x and y by elimination method 3x+3y=4. and 2x -3y=7 .

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\blue{Given}}}}}$

• Equation , ( 3x + 3y ) = 4 .....(1)
• ( 2x - 3y ) = 7 ...........(2)

$\Large{\underline{\mathfrak{\bf{\blue{Find}}}}}$

• Value of x and y , by elimination method

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

multiply by 3 in equ(2) and 2 in equ(1)

• 6x + 6y = 8
• 6x - 9y = 21

_______________Sub. its(eliminate x )

$\mapsto\sf{\:15y\:=\:-13} \\ \\ \mapsto\sf{\:y\:=\:\dfrac{-13}{15}}$

keep value of y in equ(1),

$\mapsto\sf{\:3x+3.\dfrac{-13}{15}\:=\:4} \\ \\ \mapsto\sf{\:3x\:=\:\dfrac{39}{15}+4} \\ \\ \mapsto\sf{\:3x\:=\:\dfrac{39+60}{15}} \\ \\ \mapsto\sf{\:x\:=\:\dfrac{99}{15\times 3}} \\ \\ \mapsto\sf{\:x\:=\:\dfrac{33}{15}}$

Thus:-

• Value of x = 33/5
• Value of y = - 13/15

$\Large{\underline{\underline{\mathfrak{\bf{Answer\:Verification}}}}}$.

keep value of x and y in equ(2)

$\mapsto\sf{\:(2.\dfrac{33}{15}-3.\dfrac{-13}{15})\:=\:7} \\ \\ \mapsto\sf{\:\dfrac{66}{15}-\dfrac{-39}{15}\:=\:7} \\ \\ \mapsto\sf{\:\dfrac{66+39}{15}\:=\:7} \\ \\ \mapsto\sf{\:\dfrac{\cancel{105}}{\cancel{15}}\:=\:7} \\ \\ \mapsto\sf{\:7\:=\:7} \\ \\ \mathfrak{\bf{\:\:\:\:\:\:L.H.S.\:=\:R.H.S.}}$

That's proved.

Hence, we can say that our solution is right .