Question

verify that x×(y+z)=(x×y)+(x+z) for
x=4/7,y=-3/8,z=12/-13
full solution please ​

Answer 1

Answer:

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Answer 2

★ How to do :-

Here, we are given with an equation which has some variables namely x, y and z. We are also given with the values of those variables. The variables are arranged in some ways. We are asked to verify the statement as the result that will obtain on both LHS and RHS sides of the equation should be equal. The arrangement of the variables that we can see in the question is arranged in some property. This property is called as the distributive property. In distributive property, we'll have two different types. In first type, we'll have with the multiplication and the addition and a bracket. In the second arrangement, we'll have no bracket but we'll have with the multiplication and addition. In the place of the addition sign, the subtraction sign also may be there sometimes. In this question, we are given with the addition and multiplication, so let's solve according to the addition rule. So, let's solve!!

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➤ Solution :-

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

Substitute the values of x, y and z.

${\sf \leadsto \dfrac{4}{7} \times \bigg( \dfrac{(-3)}{8} + \dfrac{12}{(-13)} \bigg)}$

First let's solve the fractions in bracket.

LCM of 8 and (-13) is 104.

${\sf \leadsto \dfrac{4}{7} \times \bigg( \dfrac{(-3) \times 13}{8 \times 12} + \dfrac{12 \times (-8)}{(-13) \times (-8)} \bigg)}$

Multiply the numerators and denominators of both fractions.

${\sf \leadsto \dfrac{4}{7} \times \bigg( \dfrac{(-39)}{104} + \dfrac{(-96)}{104} \bigg)}$

Write both numerators with a common denominator.

${\sf \leadsto \dfrac{4}{7} \times \bigg( \dfrac{(-39) + (-96)}{104} \bigg)}$

Write the second number with one sign.

${\sf \leadsto \dfrac{4}{7} \times \bigg( \dfrac{(-39) - 96}{104} \bigg)}$

Subtract the values.

${\sf \leadsto \dfrac{4}{7} \times \dfrac{(-135)}{104}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{\cancel{4} \times (-135)}{7 \times \cancel{104}} = \dfrac{1 \times (-135)}{7 \times 26}}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{(-135)}{182} \: --- LHS}$

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RHS :-

${\sf \leadsto \dfrac{4}{7} \times \dfrac{(-3)}{8} + \dfrac{4}{7} \times \dfrac{12}{(-13)}}$

Write both numerators with a common denominator.

${\sf \leadsto \dfrac{4 \times (-3)}{7 \times 8} + \dfrac{4 \times 12}{7 \times (-13)}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-12)}{56} + \dfrac{48}{(-91)}}$

Write the first fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{(-3)}{14} + \dfrac{48}{(-91)}}$

LCM of 14 and (-91) is 182.

${\sf \leadsto \dfrac{(-3) \times 13}{14 \times 13} + \dfrac{48 \times (-2)}{(-91) \times (-2)}}$

Multiply the numerators and denominators of both fractions.

${\sf \leadsto \dfrac{(-39)}{182} + \dfrac{(-96}{(-182)}}$

Write the second number with one sign.

${\sf \leadsto \dfrac{(-39 + (-96))}{182} = \dfrac{(-39) - 96}{182}}$

Subtract the numbers.

${\sf \leadsto \dfrac{(-135)}{182} \: --- RHS}$

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

Comparison :-

${\sf \leadsto \dfrac{(-135)}{182} \: and \: \dfrac{(-135)}{182}}$

As we can see that the both fractions are equal. So,

${\sf \leadsto \dfrac{(-135)}{182} = \dfrac{(-135)}{182}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!