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Question

Question: How do you solve \(4x + 8 = 7x - 7\) ?...

How do you solve 4x+8=7x74x + 8 = 7x - 7 ?

Explanation

Solution

One must remember in algebra, only the terms having the same degree or power of the variable can be added or subtracted. Here, we shift all the terms with the same power on the same side of the equation and then performing the necessary operations, the value of xx can be found.

Complete step-by-step answer:
As we know in algebra, only the terms having the same power on the variable can be added or subtracted.
Here, the terms 4  x4\;x and 7  x7\;x have the same power of xx , and hence they can be added or subtracted from each other.
Similarly, the terms   8\;8 and 7 - 7 have the same power of xx , which is   0\;0 because any number with a power of   0\;0 is equal to   1\;1 .
Hence, these terms can also be written as 8x08{x^0} and 7x0 - 7{x^0} . As they have the same power, these terms can be added or subtracted from each other.
To simplify the equation, let’s start by shifting the same power terms on the same side of the equation
The equation we are given here is
\Rightarrow 4x+8=7x74x + 8 = 7x - 7
Now, adding   7\;7 on both the sides of the equation
4x+8+7=7x7+7\Rightarrow 4x + 8 + 7 = 7x - 7 + 7
4x+15=7x+0\Rightarrow 4x + 15 = 7x + 0
Now, subtracting 4  x4\;x from both sides of the equation
4x+154x=7x4x\Rightarrow 4x + 15 - 4x = 7x - 4x
Rearranging the terms,
4x4x+15=7x4x\Rightarrow 4x - 4x + 15 = 7x - 4x
0+15=7x4x\Rightarrow 0 + 15 = 7x - 4x
Now, to explain the subtraction, let’s expand the terms 7  x7\;x and 4  x4\;x .
We know that 7  x7\;x means   7\;7 times xx . Hence, expanding 7  x7\;x as shown
7x=x+x+x+x+x+x+x\Rightarrow 7x = x + x + x + x + x + x + x
Similarly, for 4  x4\;x we can write
4x=x+x+x+x\Rightarrow 4x = x + x + x + x
Substituting these expansions in the equation,
15=(x+x+x+x+x+x+x)(x+x+x+x)\Rightarrow 15 = (x + x + x + x + x + x + x) - (x + x + x + x)
Opening the brackets,
15=x+x+x+x+x+x+xxxxx\Rightarrow 15 = x + x + x + x + x + x + x - x - x - x - x
15=x+x+x\Rightarrow 15 = x + x + x
Here we have   3\;3 times xx which can be written as 3  x3\;x ,
15=3x\Rightarrow 15 = 3x
Dividing both the sides by   3\;3 ,
153=x\Rightarrow \dfrac{{15}}{3} = x
Factoring the numerator,
5×33=x\Rightarrow \dfrac{{5 \times 3}}{3} = x

**x=5 \Rightarrow x = 5
This is the solution for the given equation. **

Note:
Another method for subtracting the terms with xx is by taking the common factor i.e. variable xx common and subtracting the numbers in the bracket. If one wants to check the accuracy of the solution, one can substitute the obtained value in the given equation and check if it satisfies the equation.