Question

Find x and y for the equations 5x-2y=11 and 3x+4y=4​

Answer 1

★ How to do :-

Here, we are given with two equations. We are asked to find the value of x and y using substitution method. By using the first equation we will find the value of x by substituting the values. Then, we use the hint of x and then we can find the value of y. Then we can find the original value of x by using the value of y. Here, we also shift numbers from one hand side to the other which changes it's sign. So, let's solve!!

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➤ Solution :-

${\sf \leadsto 5x - 2y = 11 \: --- (i)}$

${\sf \leadsto 3x + 4y = 4 \: --- (ii)}$

First let's find the value of x by using first equation.

${\tt \leadsto 5x - 2y = 11}$

Shift the number 2y from LHS to RHS, changing it's sign.

${\tt \leadsto 5x = 11 - 2y}$

Shift the number 5 from LHS to RHS.

${\tt \leadsto x = \dfrac{11 - 2y}{5}}$

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Now, let's find the value of y using the second equation.

Value of y :-

${\tt \leadsto 3x + 4y = 4}$

Substitute the value of x.

${\tt \leadsto 3 \bigg( \dfrac{11 - 2y}{5} \bigg) + 4y = 4}$

Multiply the number 3 with both numbers in brackets.

${\tt \leadsto \dfrac{33 - 6y}{5} + 4y = 4}$

Convert the number 4y to like fraction and add it with the fraction.

${\tt \leadsto \dfrac{33 - 6y + 20y}{5} = 4}$

Add the variable values on denominator.

${\tt \leadsto \dfrac{33 + 14y}{5} = 4}$

Shift the number 5 from LHS to RHS.

${\tt \leadsto 33 + 14y = 5 \times 4}$

Multiply the values on RHS.

${\tt \leadsto 33 + 14y = 20}$

Shift the number 33 from LHS to RHS, changing it's sign.

${\tt \leadsto 14y = 20 - 33}$

Subtract the values on RHS.

${\tt \leadsto 14y = (-13)}$

Shift the number 14 from LHS to RHS.

${\tt \leadsto y = \dfrac{(-13)}{14}}$

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Now, let's find the value of x by second equation.

Value of x :-

${\tt \leadsto 3x + 4y = 4}$

Substitute the value of y.

${\tt \leadsto 3x + 4 \bigg( \dfrac{(-13)}{14} \bigg) = 4}$

Multiply the number 4 with the fraction in bracket.

${\tt \leadsto 3x + \dfrac{(-52)}{14} = 4}$

Shift the fraction on LHS to RHS, changing it's sign.

${\tt \leadsto 3x = 4 - \dfrac{(-52)}{14}}$

Convert the values on LHS to like fractions.

${\tt \leadsto 3x = \dfrac{56}{14} - \dfrac{(-52)}{14}}$

Subtract those fractions now.

${\tt \leadsto 3x = \dfrac{108}{14}}$

Shift the number 3 from LHS to RHS.

${\tt \leadsto x = \dfrac{108}{14} \div \dfrac{3}{1}}$

Take the reciprocal of second fraction and multiply both fractions.

${\tt \leadsto x = \dfrac{108}{14} \times \dfrac{1}{3}}$

Write those fractions in lowest form by cancellation method.

${\tt \leadsto x = \dfrac{\cancel{108}}{14} \times \dfrac{1}{\cancel{3}} = \dfrac{36 \times 1}{14 \times 1}}$

Write the fraction in lowest form to get the answer.

${\tt \leadsto x = \cancel \dfrac{36}{14} = \dfrac{18}{7}}$

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${\red{\underline{\boxed{\bf So, \: the \: value \: of \: x \: and \: y \: is \dfrac{18}{7} \: and \: \dfrac{(-13)}{14} \: respectively.}}}}$