Chapter 9.7, Problem 15PPS

To determine

Solve the equations.

Answer to Problem 15PPS

  $\left(x,y\right)=\left(\frac{1}{2},\frac{3}{2}\right),\left(-1,6\right)$

Explanation of Solution

Given Information: $y=6x^2$ and $3x+y=3$

Formula Used: Factorization

Calculation:

  $y=6x^2$ and $3x+y=3$

So,

  $y=6x^2$ and $y=-3x+3$

Now, equate

  $6x^2=-3x+3$

So, simplify

  $6x^2+3x-3=0$

So, divide the equation by 3.

  $2x^2+x-1=0$

Now, factorize

  $2x^2+2x-x-1=0$

So,

  $2x\left(x+1\right)-1\left(x+1\right)=0$

Now,

  $\left(2x-1\right)\left(x+1\right)=0$

Hence,

  $x=\frac{1}{2}$ or $x=-1$

So, find value of $y$

  $y=6x^2$ or $y=-3x+3$

Now, put $x=\frac{1}{2}$ or $x=-1$

  $y=6{\left(\frac{1}{2}\right)}^2$ or $y=-3\times \left(-1\right)+3$

Now,

  $y=6\times \frac{1}{4}$ or $y=3+3$

Hence,

  $y=\frac{3}{2}$ or $y=6$