Question

Solve this question by substitution method
$3x + 2y = 11$
$2x + 3y = 4$

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Answer 1

Given :

• $\tt\:3x+2y=11$--(i)

• $\tt\:2x+3y=4$--(ii)

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To find :

• Value of x and y

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Concept :

In the question we are asked to find the value of x and y following the substitution method.

In order to proceed we need to know about Substitution method. In substitution method we need to acquire the value of any one variable [ say of x ] from the first equation. Then we need to substitute this value in the second equation [ value of x from (i) in equation (ii) ] . Doing so we would get the value of other variable [ y ] as well.

Hope am clear let's solve :D~

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Solution :

From equation (i) -

$\tt\:3x+2y=11$

Transposing +2y from LHS to RHS it becomes -2y

$\tt:\implies\:3x=11-2y$

Transposing 3 to RHS it goes to the denominator

$\tt:\implies\:x=\frac{11-2y}{3}$

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Now let's plug this value of x in second equation :

$\tt\:2x+3y=4$

Plugging x as $\tt\:\frac{11-2y}{3}$

$\tt:\implies\:2\times [\frac{11-2y}{3}]+3y=4$

$\tt:\implies\:\frac{22-4y}{3}+3y=4$

LCM of 1 and 3 = 3

$\tt:\implies\:\frac{(1 \times 22-4y)+(3 \times 3y)}{3}=4$

$\tt:\implies\:\frac{22-4y+9y}{3}=4$

$\tt:\implies\:\frac{22+5y}{3}=4$

Cross multiplying

$\tt:\implies\:22+5y=4 \times 3$

$\tt:\implies\:22+5y=12$

Transposing +22 from LHS to RHS it becomes -22

$\tt:\implies\:5y=12-22$

$\tt:\implies\:5y=-10$

Transposing 5 to RHS it goes to the denominator

$\tt:\implies\:y=\frac{-10}{5}$

Reducing the fraction to the lower terms

$\tt:\implies\:y=\cancel{\frac{-10}{5}}$

$\small{\underline{\boxed{\mathrm{:\implies\:y=(-2)}}}}$

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‎ Lastly let's plug the value of y in equation (i) -

$\tt\:3x+2y=11$

Plugging the value of y as ( -2 )

$\tt:\implies\:3x+2(-2)=11$

$\tt:\implies\:3x-4=11$

Transposing -4 from LHS to RHS it becomes +4

$\tt:\implies\:3x=11+4$

$\tt:\implies\:3x=15$

Transposing 3 to RHS it goes to the denominator

$\tt:\implies\:x=\frac{15}{3}$

Reducing the fraction to the lower terms

$\tt:\implies\:x=\cancel{\frac{15}{3}}$

$\small{\underline{\boxed{\mathrm{:\implies\:x=5}}}}$

$\dag\:\underline{\sf So\:the\:required\:value\:of\:x\:is\:5\:and\:y\:is\:(-2)}$

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Verification :

Plugging both the values of x as 5 and y as ( -2 ) in equation (i) -

$\tt\leadsto\:3x+2y=11$

$\tt\leadsto\:3(5)+2(-2)=11$

$\tt\leadsto\:15-4=11$

$\tt\leadsto\:11=11$

$\tt\leadsto\:LHS=RHS$

Hence Verified !~

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Therefore , ❒ x = $\large{\mathfrak\purple{5}}$

⠀ ⠀⠀ ⠀ ⠀⠀❒ y = $\large{\mathfrak\purple{(-2)}}$

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Answer 2

Refer to the attachment :))

Thnks :D``