Question

if 3a+4b is divisible by 7 then let's show that 3a^2+4b^2 is also divisible by 7 where a is not equal to b​

Answer 1

Answer:

know this might be quite trivial, but I just can't seem to figure out how to prove R={(a,b)∈Z×Z:3a+4b is divisible by 7} is a symmetric relation, ...

Answer 2

Therefore  3a²+4b² is also divisible by 7.

Step-by-step explanation:

Given that

3a + 4b is divisible by 7.

∴3a+4b=7c [ where c is a integer]

⇒3a= 7c-4b

$\Rightarrow a= \frac{7c-4b}{3}$

Now we have to show that 3a²+4b² is also divisible by 7.

∴ 3a²+4b²

$=3\times( \frac{7c-4b}{3})^2 +4b^2$    [ putting the value of a]

$=3\times( \frac{49c^2-56bc+16b^2}{9}) +4b^2$

$= \frac{49c^2-56bc+16b^2}{3} +4b^2$

$= \frac{49c^2-56bc+16b^2+12b^2}{3}$

$= \frac{49c^2-56bc+28b^2}{3}$

$=7\times ( \frac{7c^2-8bc+4b^2}{3})$ [ Taking 7 common from the numerator]

It is clearly show that 7 is a factor of  3a²+4b².

Therefore  3a²+4b² is also divisible by 7.