Question

how to solve 2x+3y-7=0,9x-2y+8=0​

Answer 1

Given :

• $\tt\:2x+3y-7=0$--(i)

• $\tt\:9x-2y+8=0$--(ii)

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To find :

• The value of x and y

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Concept :

Here we are given two linear equation. We can solve linear equation by three methods. They are :

Elimination method, Substitution method and Comparison method.

So for this question I would like to go for Elimination method.

In elimination method, we would first make sure that the leading coefficient of both the equation are same. If not we need to make it same by multiplying it by some numbers. Secondly we would subtract equation (ii) from equation (i) . Doing so we would get the value of y. Finally substituting the value of y in any of the two equation we would get the value of other variable that is x.

Hope am clear let's solve it :D~

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Solution :

Given two equations -

• Equation (i) - $\tt\:2x+3y-7=0$

• Equation (ii) - $\tt\:9x-2y+8=0$

Simplifying both the equations -

Equation (i) - $\tt\:2x+3y-7=0$

Transposing -7 to RHS it becomes +7

$\tt\:2x+3y=7$--(i)

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Equation (ii) - $\tt\:9x-2y+8=0$

Transposing +8 to RHS it becomes -8

$\tt\:9x-2y=-8$--(ii)

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Therefore , two equations are :

• $\tt\:2x+3y=7$--(i)

• $\tt\:9x-2y=-8$--(ii)

As we notice the leading coefficient are not same , multiplying equation (i) by 9 and equation (ii) by 2 .

• $\tt\:2x+3y=7$--(i) × 9

• $\tt\:9x-2y=-8$--(ii) × 2

After multiplication -

• $\tt\:18x+27y=63$--(i)

• $\tt\:18x-4y=-16$--(ii)

Subtracting equation (ii) from equation (i) -

$\sf\:Equation~(i) ~-~Equation~(ii)$

$\tt\implies\:18x+27y-(18x-4y) =63-(-16)$

Multiplying the sings and removing the brackets

$\tt\implies\:18x+27y-18x+4y =63+16$

Arranging the terms and proceeding with simple calculation

$\tt\implies\:18x-18x+27y+4y =79$

Terms with opposite signs gets cancelled

$\tt\implies\:\cancel{18x}-\cancel{18x}+27y+4y =79$

$\tt\implies\:27y+4y =79$

$\tt\implies\:31y =79$

Transposing 31 to RHS it goes to the denominator

$\small{\mathfrak{\implies\:y~=~\frac{79}{31}}}$ ★

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Substituting the value of y in equation (i) -

Equation (i) - $\tt\:2x+3y=7$

$\tt\implies\:2x+3\frac{79}{31}=7$

Multiplying the numbers

$\tt\implies\:2x+\frac{237}{31}=7$

LCM of 1 and 31 = 31

$\tt\implies\:\frac{(2x \times 31) + (237 \times 1)}{31}=7$

$\tt\implies\:\frac{62x+237}{31}=7$

Cross multiplying

$\tt\implies\:62x+237=7 \times 31$

$\tt\implies\:62x+237=217$

Transposing +237 to RHS it becomes -237

$\tt\implies\:62x=217-237$

$\tt\implies\:62x=-20$

Transposing 62 to RHS it goes to the denominator

$\tt\implies\:x=\frac{-20}{62}$

Reducing the fraction to lower terms

$\tt\implies\:x=\cancel{\frac{-20}{62}}$

$\small{\mathfrak{\implies\:x~=~\frac{-10}{31}}}$ ★

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Verifying :

Plugging both the values of x as -10/31 and y as 79/31 in equation (i) -

$\tt\:2x+3y=7$

$\tt\leadsto\:2\frac{-10}{31}+3\frac{79}{31}=7$

Multiplying the numbers

$\tt\leadsto\:\frac{-20}{31}+\frac{237}{31}=7$

LCM of 31 and 31 = 31

$\tt\leadsto\:\frac{-20+237}{31}=7$

Cross multiplying

$\tt\leadsto\:217=7 \times 31$

$\tt\leadsto\:217=217$

$\tt\leadsto\:LHS=RHS$

Hence Verified !~

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Henceforth,

‎ ⠀⠀ ⠀⠀ ⠀⠀⠀ ❒ $\small{\underline{\boxed{\mathrm\red{x~=~\frac{-10}{31}}}}}$

‎ ⠀⠀ ⠀⠀ ⠀⠀⠀ ❒ $\small{\underline{\boxed{\mathrm\red{y~=~\frac{79}{31}}}}}$

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