Question

solve by substitution method of these equations

4x - 2y = 1
3x + 2y =2

Give me the answer quickly​

Answer 1

In the above Question ,

We have to solve the following equations by substitution -

4x - 2y = 1

3x + 2y = 2 .

Solution -

Here in the first equation -

4x - 2y = 1

Now , this can be written as -

4x - 2y = 1

Adding 2y on both sides -

4x - 2y - 2y = 1 + 2y

=> 4x = 1 + 2y

Now ,

Subtracting 1 on both sides -

4x = 1 + 2y

=> 4x - 1 = 1 + 2y - 1

=> 4x - 1 = 2 y ........ { 1 }

Now,

In the second equation -

3x + 2y =2

Here , Substituting Equation 1 in place of 2 y -

=> 3x + 2y = 2

=> 3x + ( 4x - 1 ) = 2

=> 3x + 4x - 1 = 2

=> 7x - 1 = 2

=> 7x - 1 + 1 = 2 + 1 .

=> 7x = 3

=> x = { 3 / 7 }

Now ,

Substituting this in Equation 1 -

4x - 1 = 2 y

=> 4 × { 3 / 7 } - 1 = 2 y

=> { 12 / 7 } - 1 = 2 y

=> { 12 - 7 } / 7 = 2y

=> 2y = { 5 / 7 }

=> y = { 5 / 14 }

Thus, the required solution pair becomes -

x = { 7 / 3 } and y = { 5 / 14 }

_______

Answer 2

QUESTION :-

Solve by substitution method of these equations  :-

4x - 2y = 1     ...(1)

3x + 2y =2   ...(2)

SOLUTION :-

From equation (1), we get :-

$\sf{4x=1+2y}\\\\\sf{\hookrightarrow x=\dfrac{2y+1}{4} }\:\:\:\: \cdots (3)$

Putting this value of x in equation (2) :-

$\sf{\hookrightarrow 3(\dfrac{2y+1}{4})+2y=2}\\\\\sf{\hookrightarrow \dfrac{6y+3}{4}+2y=2 }\\\\\sf{\hookrightarrow \dfrac{6y+3+8y}{4} =2}\\\\\sf{\hookrightarrow 14y+3=8}\\\\\sf{\hookrightarrow 14y=5}\\\\\boxed{\boxed{\sf{\hookrightarrow y=\dfrac{5}{14} }}}$

Now, putting the value of y in equation (3) :-

$\sf{ x=\dfrac{2y+1}{4} }\\\\\sf{\hookrightarrow x=\dfrac{2(\dfrac{5}{14} )+1}{4} }\\\\\sf{\hookrightarrow x=\dfrac{\dfrac{5}{7}+1 }{4} }\\\\\sf{\hookrightarrow x=\dfrac{\dfrac{5+7}{7}}{4} }\\\\\sf{\hookrightarrow x=\dfrac{\dfrac{12}{7}}{4} }\\\\\sf{\hookrightarrow x=\dfrac{12}{7} \times \dfrac{1}{4}}\\\\\boxed{\boxed{\sf{\hookrightarrow x=\dfrac{3}{7}}}}$