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Question: How do you solve \(8x + 4\left( {4x - 3} \right) = 4\left( {6x + 4} \right) - 4\)?...

How do you solve 8x+4(4x3)=4(6x+4)48x + 4\left( {4x - 3} \right) = 4\left( {6x + 4} \right) - 4?

Explanation

Solution

In this question we have to solve the equation for xx, the given equation is a linear equation as the degree of the highest exponent of xx is equal to 1. First we have to multiply the numbers and the binomial by opening the brackets and next to solve the equation take all xx terms to one side and all constants to the other side and solve for required xx.

Complete step by step solution:
Given equation is 8x+4(4x3)=4(6x+4)48x + 4\left( {4x - 3} \right) = 4\left( {6x + 4} \right) - 4, and we have to solve forxx,
Given equation is a linear equation as the highest degree ofxxwill be equal to 1,
8x+4(4x3)=4(6x+4)4\Rightarrow 8x + 4\left( {4x - 3} \right) = 4\left( {6x + 4} \right) - 4,
Now multiply the numbers and the binomial by opening the brackets, we get,
8x+16x12=24x+164\Rightarrow 8x + 16x - 12 = 24x + 16 - 4,
Now simplifying by adding and subtracting the like terms, we get,
24x12=24x+12\Rightarrow 24x - 12 = 24x + 12,
Now transform the equation by taking all xx terms to one side and all constants to the other side we get,
24x24x=12+12\Rightarrow 24x - 24x = 12 + 12,
Now simplifying we get,
0x=24\Rightarrow 0x = 24,
By further simplifying we get,
0=240 = 24, which is not possible,
Therefore the above statement is not true, so there are no real solutions for the given equation.
Final Answer:
\therefore The value of xx when the equation 8x+4(4x3)=4(6x+4)48x + 4\left( {4x - 3} \right) = 4\left( {6x + 4} \right) - 4 will have no real solutions.

Note:
A linear equation is an equation of a straight line having a maximum of one variable. The degree of the variable will be equal to 1. To solve any equation in one variable, pit all the variable terms on the left hand side and all the numerical values on the right hand side to make the calculation solved easily.