Question

please
3x-2y=4
2x+3y=1 Elimination method ​

Answer 1

Given :

• $\sf\:3x-2y=4$ -- $\small\fbox{1}$

• $\sf\:2x+3y=1$ -- $\small\fbox{2}$

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To find :

• The value of x and y

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Solution :

Given two equations :

• $\sf\:3x-2y=4$ -- $\small\fbox{1}$

• $\sf\:2x+3y=1$ -- $\small\fbox{2}$

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As we notice the coefficient of x are not same , therefore multiplying equation 1 by 2 and equation 2 by 3 .

• $\sf\:3x-2y=4$ -- $\small\fbox{1}$ × 2

• $\sf\:2x+3y=1$ -- $\small\fbox{2}$ × 3

After multiplication -

• $\sf\:6x-4y=8$ -- $\small\fbox{3}$

• $\sf\:6x+9y=3$ -- $\small\fbox{4}$

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Subtracting equation 4 from equation 3 -

$\sf\:Equation~3~-~Equation~4~$

$\sf\implies\:6x-4y-(6x+9y)=8-3$

Multiplying the signs and removing the brackets

$\sf\implies\:6x-4y-6x-9y=8-3$

Arranging the terms

$\sf\implies\:6x-6x-4y-9y=8-3$

Number with opposite signs gets cancelled

$\sf\implies\:\cancel{6x}-\cancel{6x}-4y-9y=8-3$

$\sf\implies\:-4y-9y=8-3$

Proceeding with simple calculations

$\sf\implies\:-13y=8-3$

$\sf\implies\:-13y=5$

Transposing -13 to RHS it goes to the denominator

$\sf\implies\:y=\frac{5}{-13}$

$\small{\underline{\boxed{\mathrm{\implies\:y\:=\:(\frac{-5}{13})}}}}$ $\bf\color{red}{⋆}$

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Substituting the value of y in equation 1 -

$\sf\:3x-2y=4$ -- $\small\fbox{1}$

$\sf\implies\:3x-2(\frac{-5}{13})=4$

Multiplying the numbers

$\sf\implies\:3x-(\frac{-10}{13})=4$

Multiplying the signs and removing the brackets

$\sf\implies\:3x+\frac{10}{13}=4$

LCM of 1 and 13 = 13

$\sf\implies\:\frac{(3x \times 13) + (10 \times 1)}{13}=4$

$\sf\implies\:\frac{39x + 10}{13}=4$

Cross multiplying

$\sf\implies\:39x+10=13 \times 4$

$\sf\implies\:39x+10=52$

Transposing +10 to RHS it becomes -10

$\sf\implies\:39x=52-10$

$\sf\implies\:39x=42$

Transposing 39 to RHS it goes to the denominator

$\sf\implies\:x=\frac{42}{39}$

$\small{\underline{\boxed{\mathrm{\implies\:x\:=\:\frac{42}{39}}}}}$ $\bf\color{red}{⋆}$

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Verification :

Plugging both the values of x and y in equation 1 -

$\sf\:3x-2y=4$ -- $\small\fbox{1}$

$\sf\twoheadrightarrow\:3(\frac{42}{39})-2(\frac{-5}{13})=4$

Multiplying the numbers

$\sf\twoheadrightarrow\:\frac{126}{39}-(\frac{-10}{13})=4$

Multiplying the signs and removing brackets

$\sf\twoheadrightarrow\:\frac{126}{39}+\frac{10}{13}=4$

LCM of 39 and 13 = 39

$\sf\twoheadrightarrow\:\frac{(126 \times 1) + ( 10 \times 3)}{39}=4$

$\sf\twoheadrightarrow\:\frac{126+30}{39}=4$

Cross multiplying

$\sf\twoheadrightarrow\:126+30=4 \times 39$

$\sf\twoheadrightarrow\:156=156$

$\bf\twoheadrightarrow\:LHS~=~RHS$

Hence Verified !~

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Henceforth :

$\sf\:~~~~~~~~~~~~~$ $\bf\color{maroon}{★}$ x = $\large{\mathfrak{\frac{42}{39}}}$ ‎

$\sf\:~~~~~~~~~~~~~$ $\bf\color{maroon}{★}$ y = $\large{\mathfrak{\frac{-5}{13}}}$ ‎

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Answer 2

Step-by-step explanation:

please look into the attached image for the answer hope it will help you!