Exercise 62 Page 287

We want to solve the given system of equations using the Substitution Method. The $y\text{-}$variable is isolated in Equation (II). This allows us to substitute its value $\text{-}2x+3$ for $y$ in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the $x\text{-}$variable. We can substitute $a=1,$ $b=\text{-}1,$ and $c=\text{-}10$ into the Quadratic Formula.

$x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}$

SubstituteValues

Substitute values

$x=\frac{\text{-}(\text{-}1)\pm \sqrt{(\text{-}1{)}^2-4(1)(\text{-}10)}}{2(1)}$

▼

Solve for $x$

NegNeg

$\text{-}(\text{-}a)=a$

$x=\frac{1\pm \sqrt{(\text{-}1{)}^2-4(1)(\text{-}10)}}{2(1)}$

NegBaseToPosBase

$(\text{-}a{)}^2=a^2$

$x=\frac{1\pm \sqrt{(1{)}^2-4(1)(\text{-}10)}}{2(1)}$

BaseOne

$1^a=1$

$x=\frac{1\pm \sqrt{1-4(1)(\text{-}10)}}{2(1)}$

IdPropMult

Identity Property of Multiplication

$x=\frac{1\pm \sqrt{1-4(\text{-}10)}}{2}$

MultNegNegOnePar

$\text{-}a(\text{-}b)=a\cdot b$

$x=\frac{1\pm \sqrt{1+40}}{2}$

AddTerms

Add terms

$x=\frac{1\pm \sqrt{41}}{2}$

This result tells us that we have two solutions for $x.$ One of them will use the positive sign $x_1=\frac{1+\sqrt{41}}{2}$ and the other one will use the negative sign $x_2=\frac{1-\sqrt{41}}{2}.$ Now, consider Equation (II).

$$\begin{array}{c}y=\text{-}2x+3\end{array}$$

We can substitute the exact values of $x_1=\frac{1+\sqrt{41}}{2}$ and $x_2=\frac{1-\sqrt{41}}{2}$ into the above equation to find the values for $y.$ Let's start with $x_1.$

$y=\text{-}2x+3$

Substitute

$x=\frac{1+\sqrt{41}}{2}$

$y=\text{-}2\left(\frac{1+\sqrt{41}}{2}\right)+3$

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Solve for $y$

DenomMultFracToNumber

$2\cdot \frac{a}{2}=a$

$y=\text{-}(1+\sqrt{41})+3$

Distr

Distribute $\text{-}1$

$y=\text{-}1-\sqrt{41}+3$

AddTerms

Add terms

$y=2-\sqrt{41}$

We found that $y=2-\sqrt{41}\approx \text{-}4.4$ when $x=\frac{1+\sqrt{41}}{2}\approx 3.7.$ Therefore, one solution of the system, which is a point of intersection of the parabola and a linear function , is $\left(\frac{1+\sqrt{41}}{2},2-\sqrt{41}\right)\approx (3.7,\text{-}4.4).$ To find the other solution, we will substitute $\frac{1-\sqrt{41}}{2}$ for $x$ in Equation (II) again.

$y=\text{-}2x+3$

Substitute

$x=\frac{1-\sqrt{41}}{2}$

$y=\text{-}2\left(\frac{1-\sqrt{41}}{2}\right)+3$

▼

Solve for $y$

DenomMultFracToNumber

$2\cdot \frac{a}{2}=a$

$y=\text{-}(1-\sqrt{41})+3$

Distr

Distribute $\text{-}1$

$y=\text{-}1+\sqrt{41}+3$

AddTerms

Add terms

$y=2+\sqrt{41}$

We found that $y=2+\sqrt{41}\approx 8.4$ when $x=\frac{1-\sqrt{41}}{2}\approx \text{-}2.7.$ Therefore, our second solution is $\left(\frac{1-\sqrt{41}}{2},2+\sqrt{41}\right)\approx (-2.7,8.4).$