Question

16. Verify xx (y + 2) = x xy + x x z for:
x=3/5,y=-2/5,z=-7/5​

Answer 1

★ How to do :-

Here, we are given with some of the fractions on LHS of the statement and the same three fractions in the RHS of the statement. On LHS, the last two fractions are grouped by brackets and on the RHS, the first two fractions are grouped in the brackets. If we are given with the fractions by grouping in this format, then this property can be classified as the distributive property. This property can only be done with the multiplication and addition of fractions. It cannot be done in the subtraction and division of fractions or integers. As in this question we are given with multiplication and addition, we can verify it. We should see whether the results of LHS and RHS are equal or not. So, let's solve!!

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➤ Solution :-

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

Substitute the values of x, y and z.

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

Let's solve the LHS and RHS separately.

LHS :-

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

Write both numerators in bracket with a common denominator.

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

Write the second number in numerator with one sign.

${\sf \leadsto \dfrac{3}{5} \times \bigg( \dfrac{(-2) - 7}{5} \bigg)}$

Subtract the numerators in bracket.

${\sf \leadsto \dfrac{3}{5} \times \dfrac{(-9)}{5}}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{3 \times (-9)}{5 \times 5}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-27)}{25} \: --- LHS}$

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RHS :-

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

Write both numerators and denominators with common fraction of both brackets.

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

Multiply the numerators and denominators of both brackets.

${\sf \leadsto \dfrac{(-6)}{25} + \dfrac{(-21)}{25}}$

Write both numerators with a common denominator.

${\sf \leadsto \dfrac{(-6) + (-21)}{25}}$

Write the second number in numerator with one sign.

${\sf \leadsto \dfrac{(-6) - 21}{25}}$

Subtract the numbers.

${\sf \leadsto \dfrac{(-27)}{25} \: --- RHS}$

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Now, let's compare the results of both LHS and RHS.

Comparison :-

${\sf \leadsto \dfrac{(-27)}{25} \: and \: \dfrac{(-27)}{25}}$

As we can see that those both are equal. So,

${\sf \leadsto \dfrac{(-27)}{25} = \dfrac{(-27)}{25}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!