Chapter 9.9, Problem 3MRE

To determine

To calculate: To find the real solutions

Answer to Problem 3MRE

Solutions are $\left(3,-2\right),\left(\frac{5}{4},\frac{3}{2}\right)$

Explanation of Solution

Given information: System of equations $y^2=x+1;2x+y=4$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}{y}^2=x+1----(1)\\ 2x+y=4----(2)\end{array}$

From equation (1):

  $\begin{array}{l}{y}^2=x+1\\ x=y^2-1\end{array}$

Substituting the value in (2),

  $\begin{array}{l}2\left(y^2-1\right)+y=4\\ 2y^2-2+y=4\\ 2y^2+y-6=0\\ 2y^2+4y-3y-6=0\\ 2y\left(y+2\right)-3\left(y+2\right)=0\\ \left(y+2\right)\left(2y-3\right)=0\\ y=-2,\frac{3}{2}\end{array}$

When $y=-2$

  $\begin{array}{l}2x+\left(-2\right)=4\\ 2x=6\\ x=3\end{array}$

When $y=\frac{3}{2}$

  $\begin{array}{l}2x+\left(\frac{3}{2}\right)=4\\ 2x=4-\frac{3}{2}\\ 2x=\frac{5}{2}\\ x=\frac{5}{4}\end{array}$

Conclusion:

Hence, solutions are

  $\left(3,-2\right),\left(\frac{5}{4},\frac{3}{2}\right)$