Question

if x=1/2, y=4/3, z=-3/5 Then verify : x (y+z)=(x×y)+(x×z)​

Answer 1

Answer:

1/2(4/3+(-3/5) = (1/2×4/3) + (1/2×(-3/5)

1/2(11/15) = 2/3+(-3/5)

-11/30 = -11/30

verified

Answer 2

★ How to do :-

Here, we are given with three fractions in which we are asked to verify these three fractions in the form as given in the question. This property is also known as the distributive property. In some of the questions, we will have multiplication and addition, whereas in some of the questions, we'll have the multiplication and subtraction. In this question, we are given with the addition and multiplication. So, this can be considered as a distributive property. We should verify that the LHS and RHS of this question are equal or not. So, let's solve!!

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➤ Solution :-

${\sf \leadsto \boxed{\sf x \; (y + z)} = \boxed{\sf (x \times y) + (x \times z)}}$

Let's solve the LHS and RHS separately.

LHS :-

Substitute the values of x, y and z.

${\tt \leadsto \dfrac{1}{2} \bigg( \dfrac{4}{3} + \dfrac{(-3)}{5} \bigg)}$

First solve the fractions in bracket.

LCM of 3 and 5 is 15.

${\tt \leadsto \dfrac{1}{2} \bigg( \dfrac{4 \times 5}{3 \times 5} + \dfrac{(-3) \times 3}{5 \times 3} \bigg)}$

Multiply the numerators and denominators of both fractions in bracket.

${\tt \leadsto \dfrac{1}{2} \bigg( \dfrac{20}{15} + \dfrac{(-9)}{15} \bigg)}$

Write both numerators with a common denominator.

${\tt \leadsto \dfrac{1}{2} \bigg( \dfrac{20 + (-9)}{15} \bigg)}$

Write the second number in numerator with one sign.

${\tt \leadsto \dfrac{1}{2} \bigg( \dfrac{20 - 9}{15} \bigg)}$

Subtract the values on numerator.

${\tt \leadsto \dfrac{1}{2} \times \dfrac{11}{15}}$

Multiply the fractions now.

${\tt \leadsto \dfrac{1 \times 11}{2 \times 15} = \dfrac{11}{30} \: --- \sf LHS}$

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RHS :-

Substitute the values of x, y and z.

${\tt \leadsto \bigg( \dfrac{1}{2} \times \dfrac{4}{3} \bigg) + \bigg( \dfrac{1}{2} \times \dfrac{(-3)}{5} \bigg)}$

Write the fractions to be multiplied in a common fraction.

${\tt \leadsto \bigg( \dfrac{1 \times 4}{2 \times 3} \bigg) + \bigg( \dfrac{1 \times (-3)}{2 \times 5} \bigg)}$

Multiply the fractions given in brackets.

${\tt \leadsto \dfrac{4}{6} + \dfrac{(-3)}{10}}$

LCM of 6 and 10 is 30.

${\tt \leadsto \dfrac{4 \times 5}{6 \times 5} + \dfrac{(-3) \times 3}{10 \times 3}}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \dfrac{20}{30} + \dfrac{(-9)}{30}}$

Write both numerators with a common denominator.

${\tt \leadsto \dfrac{20 + (-9)}{30}}$

Subtract the fractions to get the value of RHS.

${\tt \leadsto \dfrac{20 - 9}{30} + \dfrac{11}{30} \: --- \sf RHS}$

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Now, let's compare the answer obtained of LHS and RHS.

${\tt \leadsto \dfrac{11}{30} \sf \: \: and \: \: \tt \dfrac{11}{30}}$

We can see that both are equal. So,

${\tt \leadsto \dfrac{11}{30} = \dfrac{11}{30}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!