Question

Solve the following simultaneous equations. 2x - 3y = 26; 2x+y=13​

Answer 1

2x - 3y = 26

2x + y = 13

(-)  (-)    (-)

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0  - 4y = 13

$-4y = 13\\y =\frac{-13}{4}$

Put this value in any one of the equation to find the value of x

2x + y = 13

2x = 13 - y

2x = $13-(\frac{-13}{4})$

2x = $13+\frac{13}{4}$

$2x=\frac{65}{4}\\\\x=\frac{65}{8}$

Answer 2

★ How to do :-

Here, we are given with two equations in which we are asked to find the values of the equations x and y. We can find the values of those variables easily by a method called as the substitution method. We use this method to find the values of the variables only when we are given with two different equations but the same variables. In those two equations the values of x will be same in both equations. The values of y will also be same to each other in both equations. But, the values of x and y may vary. We also have many other methods to find the values of the variables such as the cross multiplication method, elimination method etc... In this question, we'll solve by the substitution method. So, let's solve!!

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➤ Solution :-

${\sf \leadsto 2x - 3y = 26}$

Shift the value 3y from LHS to RHS, changing it's sign.

${\sf \leadsto 2x = 26 + 3y}$

Shift the number 2 from LHS to RHS.

${\sf \leadsto x = \dfrac{26+ 3y}{2}}$

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Value of y :-

${\sf \leadsto 2x + y = 13}$

Substitute the value of x.

${\sf \leadsto 2 \bigg( \dfrac{26 + 3y}{2} \bigg) + y = 13}$

Multiply the number 2 with the numbers in bracket.

${\sf \leadsto \dfrac{52 + 6y}{2} + y = 13}$

Write the number y in the fraction as converting it to like fraction.

${\sf \leadsto \dfrac{52 + 6y + 2y}{2} = 13}$

Add the common variables in numerator.

${\sf \leadsto \dfrac{52 + 8y}{2} = 13}$

Shift the denominator 2 from LHS to RHS.

${\sf \leadsto 52 + 8y = 13 \times 2}$

Multiply the numbers on LHS.

${\sf \leadsto 52 + 8y = 26}$

Shift the number 52 from LHS to RHS, changing it's sign.

${\sf \leadsto 8y = 27 - 52}$

Subtract the numbers on RHS.

${\sf \leadsto 8y = (-26)}$

Shift the number 8 from LHS to RHS.

${\sf \leadsto y = \dfrac{(-26)}{8}}$

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Value of x :-

${\sf \leadsto 2x - 3y = 26}$

Substitute the value of y.

${\sf \leadsto 2x - 3 \bigg( \dfrac{(-26)}{8} \bigg) = 26}$

Multiply the number 3 with all numbers in bracket.

${\sf \leadsto 2x - \dfrac{(-78)}{8} = 26}$

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

${\sf \leadsto 2x = 26 + \dfrac{(-78)}{8}}$

LCM of 8 and 1 is 8.

${\sf \leadsto 2x = \dfrac{26 \times 8}{1 \times 8} + \dfrac{(-78)}{8}}$

${\sf \leadsto 2x = \dfrac{208}{8} + \dfrac{(-78)}{8}}$

Write the second number with one sign.

${\sf \leadsto 2x = \dfrac{208 + (-78)}{8} = \dfrac{208 - 78}{8}}$

Subtract the numbers.

${\sf \leadsto 2x = \dfrac{140}{8}}$

Shift the number 2 from LHS to RHS.

${\sf \leadsto x = \dfrac{140}{8} \div 2}$

Take the reciprocal of second number and multiply both fractions.

${\sf \leadsto x = \dfrac{140}{8} \times \dfrac{1}{2}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto x = \dfrac{\cancel{140} \times 1}{\cancel{8} \times 2} = \dfrac{35 \times 1}{2 \times 2}}$

Multiply the remaining numbers.

${\sf \leadsto x = \dfrac{35}{4}}$

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${\red{\underline{\boxed{\bf So, \: the \: values \: of \: x \: and \: y \: are \: \dfrac{35}{4} \: and \: \dfrac{(-26)}{8} \: respectively.}}}}$

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