Question

How do you solve a $2a-b=ac+3b?$

Answer 1

Hint: Here as you can see that we have to solve the value of \[a\] for the expression$2a-b=ac+3b$
Collect all $'a'$ terms on the left side of the equation and remaining terms on another side. For finding the value of $'a'$ try to move $'a'$ on the left side of the equation.
On the basic concept of mathematics for solving the value of $'a'$
For example: $\left( -a+a+3a \right)$ here \[a+a\] will get cancel and equation will become $3a=4$
Here the $'3'$ division move to right side as it becomes $a=\dfrac{4}{3}$

Complete step by step solution:
As you know that given equation is,
\[\Rightarrow \]$2a-b=ac+3b$
Here for solving the value of $'a'$ collect all terms of $'a'$ on the left side and other terms on the right side.
\[\Rightarrow \]$2a-b=ac+3b$
Here for solving the value of $'a'$ collect all terms of $'a'$ on the left side and other terms on the right side.
\[\Rightarrow \]$2a-b=ac+3b$
Add $'b'$ on both sides for simplifying the equation.
\[\Rightarrow \]$2a-b+b=ac+3b+b$
Here, $-b+b$ will get cancel of left side,
\[\Rightarrow \]$2a=ac+3b+b$
Therefore, modified equation will be,
\[\Rightarrow \]$2a=ac+3b+b$
Here, $3b+1b=4b$
So,
\[\Rightarrow \]$2a=ac+4b$
Subtract $'ac'$ from both sides,
\[\Rightarrow \]$2a-ac=ac-ac+4b$
As in above equation $'+ac-ac'$ will get cancel,
\[\Rightarrow \]$2a-ac=4b$
Here, on the left side, take $'a'$ as a common factor or term.
\[\Rightarrow \]$a\left( 2-c \right)=4b$
As, to simplify the above equation divide both sides by $\left( 2-c \right)$
\[\Rightarrow \]$\dfrac{a\left( 2-c \right)}{\left( 2-c \right)}=\dfrac{4b}{\left( 2-c \right)}$
$\left( 2-c \right)$ in numerator and denominator gets canceled.
\[\Rightarrow \]$\dfrac{a\left( 2-c \right)}{\left( 2-c \right)}=\dfrac{4b}{\left( 2-c \right)}$
Hence,
\[\Rightarrow \]$a=\dfrac{4b}{2-c}$

The value of $a$ is $\dfrac{4b}{2-c}$

Additional Information:
In algebra, a quadratic equation (From the Latin quadrates for ‘square’) is any equation that can be rearranged in standard form as $a{{x}^{2}}+bx+c=0$
_{\[\Rightarrow \]$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$}
The quadratic formula for the roots of the general quadratic equation.
Where, $x$represents on unknown, and $a,b$ and $c$ represent known numbers, where $a\ne 0.$ If $a=0$ then the equation is linear not quadratic as there is no $a{{x}^{2}}$ term. The numbers $a,b$ and $c$ are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.

Note: As you have to solve the value of $'a'$ move it on the left side. Use the basic concept of mathematics to find the value of $'a'$
For finding the value of $'a'$ here you have used different arithmetic operations such as addition, subtraction, multiplication and division. You can also take common factors which will help to save the time.