Precalculus

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 11, Problem 69RE

To determine

Solve each system of equations.

Expert Solution & Answer

Answer to Problem 69RE

$(−2,1)and(−25,115)$

Explanation of Solution

Given information:

Solve each system of equations.

${2x+y+3=0x2+y2=5$

Calculation:

We can use the method of substitution for solving this system of equations.

Add $−y−3$ to both sides of the equation $2x+y+3=0$ .

$2x+y+3−3−2x=0−3−2x$

$y=−3−2x$

Substitute $−3−2x$ for $y$ in the equation $x2+y2=5$ and simplify.

$x2+(−3−2x)2=5$

$x2+9+4x2+12x=5$

$5x2+12x+9=5$

Subtract $5$ from both sides of the equation.

$5x212x+9−5=5−5$

$5x212x+4=0$

This can be solved by using the quadratic formula $x=−b±b2−4ac2a$ where $a=5,b=12,c=4$ .

$x=−12±122−4.5.42.5$

$=−12±144−8010$

$=−12±6410$

$=−12±810$

Simplify $x=−12±810$ to get values for $y$ .

$x=−12±810$ Or $−12−810$

$=−410$ Or $−2010$

$=−25$ Or $−2$

Substitute $=−25$ for $x$ in the equation $y=−3−2x$ to find $y$ .

$y=−3−2(−25)$

$=−3+45$

$=−15+45$

$=−115$

Now, substitute $−2$ for $x$ in the equation $y=−3−2x$ to find $y$ .

$y=−3−2(−2)$

$=−3+4$

$=1$

Hence, the solutions, expressed on ordered pairs, are $(−2,1)$ and $(−25,115)$

Chapter 11, Problem 68RE

Chapter 11, Problem 70RE

Chapter 11 Solutions

Precalculus

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