Question

Subtract the sum of -3a(a + b + c) and 2a(a-b+c) from the difference of a(a-3c) and 7a(b + c).​

Answer 1

We have to subtract the sum of$-3a(a + b + c)$ and$2a(a-b+c)$ from the difference of$a(a-3c) \ and\ 7a(b + c).$

• By using the Bodmas rule, we are solving the above equation.
• As we know that the Bodmas rule is used to remember the order of operations to be followed while solving expressions in mathematics.

Where,

$\begin{array}{l}\mathrm{B}=\text{brackets}\\\mathrm{O}=\text { order of powers or rules } \\\mathrm{D}=\text { division } \\\mathrm{M}=\text { multiplication } \\\mathrm{A}=\text { addition } \\\mathrm{S}=\text { subtraction }\end{array}$

We are solving in the following way:

We have,

The sum of$-3a(a + b + c)$ and$2a(a-b+c)$ will be:

$\Rightarrow -3a(a + b + c) + 2a(a-b+c)\\\Rightarrow -3a^{2}-3ab-3ac+2a^{2}-2ab+2ac\\\Rightarrow-1 a^{2}-5ab-1ac$

Hence, the sum of$-3a(a + b + c)$ and$2a(a-b+c)$ is$-1 a^{2}-5ab-1ac$.

Now, the difference of $a(a-3c) \ and\ 7a(b + c)$ will be:

$a(a-3c) - 7a(b + c).\\\Rightarrow a^{2} -3ac-7ab-7ac$

Hence, the difference of $a(a-3c) \ and\ 7a(b + c)$is$a^{2} -3ac-7ab-7ac$  

Now,

Subtracting the sum of$-3a(a + b + c)$ and$2a(a-b+c)$ from the difference of$a(a-3c) \ and\ 7a(b + c).$

$(-1 a^{2}-5ab-1ac)$$-(a^{2} -3ac-7ab-7ac)$

$\Rightarrow -1a^{2} -5ab-1ac-a^{2} +3ac+7ab+7ac\\\Rightarrow -2a^{2} +2ab+9ac$

Hence, the solution of the above equation is$-2a^{2} +2ab+9ac$.