Question

How do you solve $y = 3{x^2} - x - 2,y = - 2x + 2$ using substitution?

Answer 1

Hint: Since we have the second equation as a linear equation, we will put the value of y in the given first equation, then we will obtain a quadratic in x and solve it. Now, to find y, we will put that x in the second equation.

Complete step-by-step solution:
We are given that we are required to solve $y = 3{x^2} - x - 2,y = - 2x + 2$ using substitution.
Let us terms the given equation $y = 3{x^2} - x - 2$ as equation number 1 and the given equation $y = - 2x + 2$ as equation number 2.
We can now put in equation number 2 in equation number 1.
We will then obtain the following equation:-
$ \Rightarrow - 2x + 2 = 3{x^2} - x - 2$
Re – arranging the terms, we will then obtain the following equation:-
$ \Rightarrow 3{x^2} - x - 2 + 2x - 2 = 0$
Now, we will club the constant terms and the terms with x, we will then obtain the following equation:-
$ \Rightarrow 3{x^2} + x - 4 = 0$
Now, we will split the middle term, and then we can rewrite the above equation as follows:-
$ \Rightarrow 3{x^2} - 3x + 4x - 4 = 0$
Now, we will take 3x common from the first two terms and we will then obtain the following equation:-
$ \Rightarrow 3x(x - 1) + 4x - 4 = 0$
Taking out 4 common from the last two terms in the above equation, we will then obtain the following equation:-
$ \Rightarrow 3x(x - 1) + 4(x - 1) = 0$
Now, taking out (x – 1) as the common factor from the above expression, we will then obtain the following equation:-
$ \Rightarrow (x - 1)(3x + 4) = 0$
Now, we see that either x = 1 or $x = - \dfrac{4}{3}$.
Putting this in equation number 2, we will then obtain the following values of y:-
Either $y = - 2 + 2 = 0$ or $y = - 2\left( { - \dfrac{4}{3}} \right) + 2 = \dfrac{{14}}{3}$.

Thus, x = 1 or $- \dfrac{4}{3}$ and y= 0 or $\dfrac{{14}}{3}$ is the required answer.

Note: The students must note that you may use alternate methods for solving the equations other than using the method of “splitting the middle term”.
Alternate Way :
$ \Rightarrow 3{x^2} + x - 4 = 0$
The general quadratic equation is given by $a{x^2} + bx + c = 0$ and its roots are given by:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing it to the given equation, we have a = 3, b = 1 and c = - 4.
So, the roots of the equation are:-
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(3)( - 4)} }}{{2(3)}}$
Simplifying the calculations in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 48} }}{6}$
Simplifying the calculations further in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{ - 1 \pm 7}}{6}$
Thus, the roots are $1, - \dfrac{4}{3}$.