Question

How do you solve $\left( {x + 5} \right) - 2\left( {4x - 1} \right) = 0$?

Answer 1

Hint: Here we just need to open the bracket and simplify multiply the terms and add and subtract the terms in order to get the value of $x$ and here we must know that $a\left( {b + c} \right) = ab + ac$ and if we take one term from LHS to RHS or vice versa we need to change its sign like if we have $a + c = b$ then we can also write it as $a = b - c$.

Complete step by step solution:
Here we are given to solve the function given as $\left( {x + 5} \right) - 2\left( {4x - 1} \right) = 0$
Solving means to get the value of the unknown variable which is $x$ over here. So we can write the terms containing $x$ on one side and all the constants on the other side. First we need to know that
$a\left( {b + c} \right) = ab + ac$.
Applying this property to open the brackets in the given function we will get:
$\left( {x + 5} \right) - 2\left( {4x - 1} \right) = 0$
$x + 5 - 2\left( {4x} \right) - 2\left( { - 1} \right) = 0$
On further simplification we get:
$x + 5 - 8x + 2 = 0$
So we can add the terms with the coefficient of $x$ and the constants separately and we will get:
$
  x + 5 - 8x + 2 = 0 \\
   - 7x + 7 = 0 \\
 $
Now we can take $ + 7$ to RHS and we can change the sign from plus to minus and we will get:
$
   - 7x + 7 = 0 \\
   - 7x = - 7 \\
 $

Hence we can get $x = \dfrac{{ - 7}}{{ - 7}} = 1$

Here we have just applied the property by which we must know that if two terms are multiplying and we take one term to the other side like from LHS to RHS then the sign from multiplication will change to division.

Note:
In these types of problems, the student must know that this is the linear equation in $x$ as the degree of $x$ is $1$ but in the quadratic equation the degree is $2$ hence we will get the two values of $x$ and it can also be possible that both are same or different.