Question

$\frac{2x+3}{x-3}$=$\frac{3}{5}$

Answer 1

Answer:

QUESTION:

$\frac{2x+3}{x-3}$=$\frac{3}{5}$

SOLUTION:

$5(2x + 3) = 3(x - 3)$

$10x + 15 = 3x - 3$

$10x - 3x = - 3 - 15$

$7x = - 18$

$x = \frac {-18}{7}$

Answer 2

★ How to do :-

Here, we are given with an equation in which we are given with two same variables x in numerator and denominator each. We are also given with an equivalent fraction on the RHS side of the equation. We are asked to find the value of x. To find the value of x, we use an other concept which is known as the cross multiplication method. This method. an only be used for fractions. This method is used to find the value of x in the fractions and to compare the fractions. In this question, we use it to find the value of x. We can also verify our answer at last by substituting the value of x. So, let's solve!!

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➤ Solution :-

${\sf \leadsto \dfrac{2x + 3}{x - 3} = \dfrac{3}{5}}$

Cross multiply the numbers.

${\sf \leadsto 5 (2x + 3) = 3 (x - 3)}$

Multiply the number outside the bracket with the numbers in the bracket.

${\sf \leadsto 10x + 15 = 3x - 9}$

Shift all variables on LHS and constants on RHS.

${\sf \leadsto 10x - 3x = (-9) - 15}$

Subtract the values on LHS and RHS.

${\sf \leadsto 7x = (-24)}$

Shift the number 7 from LHS to RHS.

${\sf \leadsto x = \dfrac{(-24)}{7}}$

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${\red{\underline{\boxed{\bf So, \: the \: value \: of \: x \: is \: \dfrac{(-24)}{7}.}}}}$

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Verification :-

${\sf \leadsto \dfrac{2x + 3}{x - 3} = \dfrac{3}{5}}$

Substitute the value of the x.

${\sf \leadsto \dfrac{2 \times \dfrac{(-24)}{7} + 3}{\dfrac{(-24)}{7} - 3} = \dfrac{3}{5}}$

Multiply the number 2 with the fraction on numerator in LHS.

${\sf \leadsto \dfrac{\dfrac{(-48)}{7} + 3}{\dfrac{(-24)}{7} - 3} = \dfrac{3}{5}}$

Convert the unlike fractions to like fractions on both numerator and denominator.

${\sf \leadsto \dfrac{\dfrac{(-48) + 21}{7}}{\dfrac{(-24) - 21}{7}} = \dfrac{3}{5}}$

Add and subtract the fractions on numerator and denominator.

${\sf \leadsto \dfrac{\dfrac{(-27)}{7}}{\dfrac{(-45)}{7}} = \dfrac{3}{5}}$

Cancel the number 7 on numerator and denominator.

${\sf \leadsto \dfrac{(-27)}{(-45)} = \dfrac{3}{5}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{3}{5} = \dfrac{3}{5}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!