Question
Question: How do you solve \(7 - 7x = (4x + 9)(x - 1)\)?...
How do you solve 7−7x=(4x+9)(x−1)?
Solution
In this question, we are asked to find the value of the variable x. Firstly, we multiply the terms in the right hand side with each other by using distributive property and simplify it. Then we transfer all the terms to the left hand side and simplify it by combining the like terms. After that we obtain a quadratic equation, which we simplify using the factoring method. By the obtained factors we find the value of the unknown variable x.
Complete step by step answer:
Given an equation of the form,
7−7x=(4x+9)(x−1) …… (1)
We are asked to find the solution of the above equation (1). We do this by finding the value of the unknown variable x.
Now we try to simplify the given equation.
Firstly consider the L.H.S. given by (4x+9)(x−1).
We now simplify this using the distributive property of subtraction, which is given by a(b−c)=a⋅b−a⋅c
Here a=4x+9, b=x and c=1.
So we get,
⇒(4x+9)(x−1)=(4x+9)(x)−(4x+9)(1)
Now again applying the distributive property of addition which is given by
(b+c)a=b⋅a+c⋅a
So applying to each of the pairs in the R.H.S. we get,
⇒(4x⋅x+9⋅x)−(4x⋅1+9⋅1)
Now removing the parentheses we get,
⇒4x2+9x−4x−9
Substituting this in the equation (1) we get,
⇒7−7x=4x2+9x−4x−9
Now transfer all the terms on the L.H.S. to the R.H.S. we get,
⇒0=4x2+9x−4x−9−7+7x
Rearrange the terms we get,
⇒0=4x2+9x−4x+7x−9−7
Combining the like terms 9x−4x+7x=12x and −9−7=−16
Hence we obtain,
⇒0=4x2+12x−16
Hence we have the equation 4x2+12x−16=0 …… (2)
Note that this is a quadratic equation and we solve this using the factoring method.
The equation (2) can also be written as,
4x2+16x−4x−16=0
Now factor out the common terms we get,
⇒4x(x+4)−4(x+4)=0
⇒(x+4)(4x−4)=0
Hence we have either x+4=0 or 4x−4=0.
If x+4=0, we have x=−4.
If 4x−4=0, we have
⇒4x=4
Now dividing throughout by 4 we get,
⇒44x=44
⇒x=1
Hence the solution for the equation 7−7x=(4x+9)(x−1) is x=−4 and x=1.
Note: In such types of questions students mainly don’t get an approach how to solve the given equation. In such types of questions students mainly forget to use the correct operations on the equation such that the expression gets simplified.
We can check whether the obtained values are correct, by substituting back it to the given equation.
If the values on L.H.S. and R.H.S. come to be the same, then our solution is correct.