Int-Alg Chapter 9 Equations and Graphs

Section A.9 Chapter 9 Equations and Graphs

Subsection A.9.1 Properties of Lines

Subsubsection A.9.1.1 Find the slope of a line

We can tell whether two lines are parallel, perpendicular, or neither by comparing their slopes.

Example A.9.1.

Find the slope of the line $~6x-8y = 9$

Solution.

The easiest way to find the slope of this line is to put its equation into slope-intercept form by solving for $y\text{.}$

\begin{align*} -8y \amp = -6x+9\\ y \amp = \dfrac{-6x+9}{-8} = \dfrac{3}{4}x - \dfrac{9}{8} \end{align*}

The slope of the line is $\dfrac{3}{4}\text{.}$

Example A.9.2.

Find the slope of the line whose intercepts are $(112,0)$ and $(0,140)\text{.}$

Solution.

We use the slope formula.

\begin{equation*} m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{140-0}{0-112} = \dfrac{140}{-112} = \dfrac{-5}{4} \end{equation*}

Checkpoint A.9.3.

Find the slope of the line $~2.8x+3.6y = 1.2$

Answer.

$\dfrac{7}{9}$

Checkpoint A.9.4.

Find the slope of the line $~6x=72$

Answer.

undefined

Checkpoint A.9.5.

Find the slope of the line that passes through $(15,-6)$ and $(10,-3)\text{.}$

Answer.

$\dfrac{-3}{5}$

Checkpoint A.9.6.

Find the slope of a line that describes a 35% grade (or incline).

Answer.

$\dfrac{7}{20}$

Subsubsection A.9.1.2 Use the point-slope formula

Recall the point-slope formula for finding the equation of a line:

\begin{equation*} y-y_1 = m(x-x_1) \end{equation*}

Example A.9.7.

Find the $y$-intercept of the line of slope $\dfrac{-2}{3}$ that passes through $(-8,9)\text{.}$

Solution.

We first use the point-slope formula to find the equation of the line.

\begin{align*} y-9 \amp = \dfrac{-2}{3}(x+8) \amp \amp \blert{\text{Multiply by 3 to clear the fraction.}}\\ 3y-27 \amp = -2x-16 \amp \amp \blert{\text{Add 27 to both sides.}}\\ 3y \amp = -2x+11 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y \amp = \dfrac{-2}{3}x + \dfrac{11}{3} \end{align*}

The $y$-intercept of the line is $\left(0, \dfrac{11}{3}\right)\text{.}$

Example A.9.8.

Find the equation of the line with $x$-intercept $(2,0)$ that passes through $(-3,-3)\text{.}$

Solution.

We first compute the slope, using the two points given.

\begin{equation*} m = \dfrac{0+3}{2+3} = \dfrac{3}{5} \end{equation*}

Now we can use the point-slope formula.

\begin{align*} y-0 \amp = \dfrac{3}{5}(x-2) \amp \amp \blert{\text{Multiply by 5 to clear the fraction.}}\\ 5y \amp = 3x-6 \amp \amp \blert{\text{Divide both sides by 5.}}\\ y \amp = \dfrac{3}{5}x - \dfrac{6}{5} \end{align*}

Checkpoint A.9.9.

Find the equation of the line that has slope $\dfrac{-1}{3}$ and passes through $(-4,-6)\text{.}$

Answer.

$y = \dfrac{1}{3}x=\dfrac{22}{3}$

Checkpoint A.9.10.

Find the equation of the line that has slope $0$ and passes through $(25,64)\text{.}$

Answer.

$y=64$

Checkpoint A.9.11.

Find the $y$-intercept of the line that passes through $(-5,49)$ and has slope $\dfrac{-4}{5}\text{.}$

Answer.

$(0,0)$

Checkpoint A.9.12.

Find the $y$-intercept of the line that has slope $\dfrac{-7}{3}$ and passes through $(-6,0)\text{.}$

Answer.

$(0,-14)$

Subsubsection A.9.1.3 Use properties of geometric figures

Analytic geometry uses algebra to help solve geometric problems.

Example A.9.13.

The figure shows isosceles triangle $ABC$ and its altitude $\overline{AP}\text{.}$ (Recall that the altitude of a triangle is the segment perpendicular to the base that passes through the opposite vertex.) For this problem, we’ll use the following property:

In an isoceles triangle, the altitude bisects the base.

$isosceles triangle$

Find the equation of the line that includes $\overline{AP}\text{.}$
Find the coordinates of point $P\text{.}$
Find the length of the segment $\overline{AP}\text{.}$

Solution.

Because $\overline{AP}$ is perpendicular to $\overline{BC}\text{,}$ we can find its slope. The coordinates of $B$ and $C$ are $(6,0)$ and $(2,-6)\text{,}$ so the slope of $\overline{BC}$ is

\begin{equation*} \dfrac{-6-0}{-2-6} = \dfrac{3}{4} \end{equation*}

The slope of $\overline{AP}$ is the negative reciprocal of $\dfrac{3}{4}\text{,}$ or $\dfrac{-4}{3}\text{.}$

Now we can use the point-slope formula with $m=\dfrac{-4}{3}$ and the coordinates of $A(-4,5)$ to calculate the equation of the line.

\begin{align*} \dfrac{-4}{3} \amp = \dfrac{y-5}{x+4} \amp \amp \blert{\text{Cross-multiply.}}\\ 3(y-5) \amp = -4(x+4) \amp \amp \blert{\text{Apply the distributive law.}}\\ 3y-15 \amp = -4x-16 \amp \amp \blert{\text{Add 15 to both sides.}}\\ 3y \amp = -4x-1 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y \amp = \dfrac{-4}{3}x - \dfrac{1}{3} \end{align*}
Because the altitude bisects the base, point $P$ is the midpoint of $\overline{BC}\text{.}$ The coordinates of $B$ and $C$ are $(6,0)$ and $(2,-6)\text{,}$ so we use the midpoint formula to find the coordinates of $P\text{.}$

\begin{align*} x \amp = \dfrac{-2+6}{2} = 2\\ y \amp = \dfrac{-6+0}{2} = -3 \end{align*}

The coordinates of $P$ are $(2,-3)\text{.}$
The coordinates of $P$ are $(2,-3)\text{,}$ and the coordinates of $A$ are $(-4,5)\text{.}$ We use the distance formula to find the length of $\overline{AP}\text{.}$

\begin{equation*} \overline{AP} = \sqrt{(-4-2)^2+(5+3)^2} = \sqrt{36+64} = 10 \end{equation*}

Checkpoint A.9.14.

The line $y=-2x+9$ is tangent at point $Q$ to a circle with center $C(-4,2)\text{.}$ For this problem, use the following property:

The tangent to a circle is perpendicular to the radius through the point of tangency.

$circle and tangent line$

Find the equation of the line through $C$ and $Q\text{.}$
Find the coordinates of $Q\text{.}$
Find the radius of the circle, $r=CQ\text{.}$
Find the equation of the circle.

Answer.

$\displaystyle y=\dfrac{1}{2}x=4$
$\displaystyle (2,5)$
$\displaystyle 3\sqrt{5}$
$\displaystyle (x+4)^2+(y-2)^2=45$

Checkpoint A.9.15.

In the right triangle shown, $\overline{ST}$ is parallel to the shorter leg, and $h=\dfrac{9}{10}r\text{.}$ For this problem, use the following property:

A line parallel to the base of a triangle cuts off a similar triangle. (Recall that the sides of similar triangles are proportional.)

$triangle cut parallel to base$

Write an expression for $r$ in terms of $h\text{.}$
Find an expression for $b$ in terms of $h\text{.}$
If $OS=\dfrac{2}{3}h$ find an expression for $ST$ in terms of $h\text{.}$

Answer.

$\displaystyle r=\dfrac{10}{9}h$
$\displaystyle b=\dfrac{\sqrt{19}}{9}h$
$\displaystyle ST = \dfrac{2\sqrt{19}}{27}h$

Checkpoint A.9.16.

The quadrilateral $ABCD$ has vertices $~A(-4,-1),~B(-2,5),~C(8,5)~$ and $~D(6,-1).$

$triangle cut parallel to base$

Show that $ABCD$ is a parallelogram (its opposite sides are parallel).
Find equations for the lines through the diagonals, $\overline{AC}$ and $\overline{BD}\text{.}$
Find the intersubsection of the diagonals, $P\text{.}$
Find the lengths of $\overline{AP}$ and $\overline{PC}\text{,}$ and the lengths of $\overline{BP}$ and $\overline{PD}\text{.}$
This example illustrates the following property of parallelograms:

The diagonals of a parallelogram each other.

Answer.

$\displaystyle m_{\overline{AB}}=m_{\overline{CD}}=\dfrac{3}{2}; ~m_{\overline{AD}}=m_{\overline{BC}}=0$
$\displaystyle y=\dfrac{1}{2}x+1; ~y=\dfrac{-3}{4}x+\dfrac{7}{2}$
$\displaystyle (2,2)$
$\displaystyle AP=PC=3\sqrt{5}; ~BP=PD=5$
bisect

Subsection A.9.2 The Distance and Midpoint Formulas

Subsubsection A.9.2.1 Use radicals

Because the distance formula is derived from the Pythagorean theorem, using it involves working with radicals.

Example A.9.17.

In the right triangle shown, $a=8$ and $b=6\text{.}$

Is $c=a+b\text{?}$ Why or why not?
Find the length of $c\text{.}$

$right triangle$

Solution.

No, the length of $c$ is shorter than the lengths of $a$ and $b$ combined, so $c \lt a+b\text{.}$
We use the Pythagorean theorem:

\begin{align*} c^2 \amp = a^2+b^2 \amp \amp \blert{\text{Substitute }a=8~ \text{and }b=6~ \text{and evaluate.}}\\ c^2 \amp = 8^2+6^2=64+36=100 \amp \amp \blert{\text{Take square roots.}}\\ c \amp = \sqrt{100} = 10 \end{align*}

The hypotenuse is 10 cm long.

Example A.9.18.

Is $\sqrt{a^2+b^2} = a+b\text{?}$

Solution.

No. In the previous example, $c=\sqrt{a^2+b^2}\text{,}$ and we saw that $c \lt a+b\text{.}$ We cannot simplify $\sqrt{a^2+b^2}$ by taking the square root of each term.

Example A.9.19.

If $r=\sqrt{14}\text{,}$ write and simplify expressions for:

$\displaystyle r \cdot r$
$\displaystyle r+r$

Solution.

$r \cdot r = \left(\sqrt{14}\right)\left(\sqrt{14}\right) = 14\text{.}$ Or, $r \cdot r = r^2 = \left(\sqrt{14}\right)^2 = 14\text{.}$
$r+r = \sqrt{14} + \sqrt{14} = 2\sqrt{14}\text{.}$ Or, $r+r = 2r = 2\sqrt{14}\text{.}$

Checkpoint A.9.20.

Simplify if possible: $~\sqrt{(x-4)^2+(y-2)^2}$

Answer.

cannot be simplified

Checkpoint A.9.21.

Simplify if possible: $~\sqrt{7^2+w^2}$

Answer.

$\sqrt{49+w^2}$

Checkpoint A.9.22.

Simplify if possible: $~\sqrt{7+w} \sqrt{7+w}$

Answer.

$7+w$

Subsubsection A.9.2.2 Use the equation for a circle

The equation for a circle of radius $r$ centered at $(h,k)$ is

\begin{equation*} \blert{(x-h)^2+(y-k)^2=r^2} \end{equation*}

Example A.9.23.

Does $(-1,5)$ lie on the circle $~(x-1)^2+(y-3)^2=8~\text{?}$

Solution.

We substitute $x=\alert{-1},~ y=\alert{5}$ into the equation for the circle.

\begin{equation*} (\alert{-1}-1)^2+(\alert{5}-3)^2 = (-2)^2+2^2 = 4+4 = 8 \end{equation*}

The point $(-1,5)$ satisfies the equation, so it does lie on the circle.

Checkpoint A.9.24.

Find a point with $y$-coordinate $-2$ that lies on the circle $(x-2)^2+(y+1)^2=9\text{.}$

Answer.

$(2+\sqrt{8}, -2)$ and $(2-\sqrt{8}, -2)$

Subsubsection A.9.2.3 Complete the square

To find the center and radius of a circle, we may need to complete the square.

Example A.9.25.

Write an equivalent equation in which the left side is a perfect square: $~x^2-6x=2$

Solution.

We want to find a constant $p^2$ so that $x^2\blert{-6x}+p^2$ is a perfect square, namely $(x+p)^2\text{.}$ Now, $~(x+p)^2 = x^2\blert{+2px}+p^2~\text{,}$ so $2p=-6\text{,}$ and $p=-3\text{.}$ Thus, we add $p^2=\alert{9}$ to both sides of the equation.

\begin{equation*} x^2-6x+\alert{9} = 2+\alert{9} \end{equation*}

Now we can write the left side as a perfect square:

\begin{equation*} (x-3)^2 = 11 \end{equation*}

Example A.9.26.

Write an equivalent equation in which the left side is a perfect square: $~y^2+7y=5$

Solution.

For this equation, $2p=7$ so $p = \dfrac{7}{2}\text{.}$ We add $p^2=\left(\dfrac{7}{2}\right)^2=\alert{\dfrac{49}{4}}$ to both sides of the equation.

\begin{equation*} y^2+7y+\alert{\dfrac{49}{4}}=5+\alert{\dfrac{49}{4}} \end{equation*}

We write the left sides as a perfect square, $(x+p)^2\text{,}$ and simplify the right side.

\begin{equation*} \left(x+\dfrac{7}{2}\right)^2=\dfrac{69}{4}~~~~~~~~~~~~\blert{5+\dfrac{49}{4}=\dfrac{20}{4}+\dfrac{49}{4}=\dfrac{69}{4}} \end{equation*}

Checkpoint A.9.27.

Write an equivalent equation in which the left side is a perfect square: $~x^2+12x=-6$

Answer.

$(x+6)^2=30$

Checkpoint A.9.28.

Write an equivalent equation in which the left side is a perfect square: $~y^2-5y=3$

Answer.

$\left(y-\dfrac{5}{2}\right)^2=\dfrac{37}{4}$

Subsection A.9.3 Conic Sections: Ellipses

Subsubsection A.9.3.1 Find points on a graph

Points on the graph of a conic subsection satisfy a quadratic equation in two variables.

Example A.9.29.

Find all points on the graph of $x^2+y^2 = 12$ with $x$-coordinate 2.

Solution.

We substitute $x=2$ into the equation to obtain

\begin{equation*} 2^2+y^2=12 \end{equation*}

and simplify to $y^2=8\text{.}$ Solving for $y\text{,}$ we find $y=\pm \sqrt{8} = \pm 2\sqrt{2}\text{.}$ Thus, the points on the graph of $x^2+y^2 = 12$ with $x$-coordinate 2 are $(2, 2\sqrt{2})$ and $(2, -2\sqrt{2})\text{.}$

Checkpoint A.9.30.

Find all points on the graph of $y=2x^2-4x+3$ with $y$-coordinate 3.

Answer.

$(0,3),~(2,3)$

Checkpoint A.9.31.

Find all points on the graph of $(x-4)^2+(y+1)^2=25$ with $y$-coordinate 3.

Answer.

$(1,3)~(7,3)$

Checkpoint A.9.32.

Find all points on the graph of $\dfrac{(x-2)^2}{16}+\dfrac{(y+1)^2}{9} = 1$ with $x$-coordinate $-1\text{.}$

Answer.

$\left(-1,-1+\dfrac{3\sqrt{7}}{4}\right),~\left(-1,-1-\dfrac{3\sqrt{7}}{4}\right)$

Subsubsection A.9.3.2 Complete the square

To graph a conic subsection, we complete the square in each variable to put the equation in standard form.

Example A.9.33.

Solve $~2x^2+3 = 8x~$ by completing the square.

Solution.

We begin by isolating the constant term on the right side of the equation.

\begin{align*} 2x^2 - 8x \amp =-3 \amp \amp \blert{\text{Factor out the coefficient of} ~x^2.}\\ 2(x+2-4x~~~~~~ \amp = -3) \amp \amp \blert{\text{Complete the square.}}\\ 2(x^2-4x \blert{+ 4}) \amp = -3 \blert{+ 4} \amp \amp \blert{\text{Simplify each side.}}\\ 2(x-2)^2 \amp = 5 \amp \amp \blert{\text{Isolate the perfect square.}}\\ (x-2)^2 \amp = \dfrac{5}{2} \amp \amp \blert{\text{Extract roots.}}\\ x-2 \amp = \pm \sqrt{\dfrac{5}{2}} \amp \amp \blert{\text{Sovle for}~x.}\\ x \amp = 2 \pm \sqrt{\dfrac{5}{2}} \end{align*}

Checkpoint A.9.34.

Solve $~x^2-7x=4~$ by completing the square.

Answer.

$\dfrac{7 \pm \sqrt{65}}{2}$

Checkpoint A.9.35.

Solve $~3x^2+6x-2=0~$ by completing the square.

Answer.

$-1 \pm \sqrt{\dfrac{5}{3}}$

Checkpoint A.9.36.

Solve $~2x^2-8=3x~$ by completing the square.

Answer.

$\dfrac{3 \pm \sqrt{73}}{4}$

Subsection A.9.4 Conic Sections: Hyperbolas

Subsubsection A.9.4.1 Write a quadratic equation in standard form

The parameters in the standard form determine the shape of the graph.

Example A.9.37.

Write the equation $~4x^2-3y^2=1~$ in the form $~\dfrac{x^2}{A}-\dfrac{y^2}{B} = 1$

Solution.

Recall that dividing by a fraction is equivalent to multiplying by its reciprocal. For instance, $~\dfrac{x^2}{\dfrac{1}{2}} = 2x~\text{,}$ and $~\dfrac{a}{\dfrac{3}{4}} = \dfrac{4}{3}a~$

Thus, for our example,

\begin{equation*} ~4x^2 = \dfrac{x^2}{\dfrac{1}{4}},~~~ \text{and} ~~~3y^2 = \dfrac{y^2}{\dfrac{1}{3}}~ \end{equation*}

So we can write the equation as $~\dfrac{x^2}{\dfrac{1}{4}} - \dfrac{y^2}{\dfrac{1}{3}} = 1\text{.}$

Write each equation in the form $~\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$

Checkpoint A.9.38.

$\dfrac{4}{9}(x-2)^2 - \dfrac{1}{5}(y+2)^2 = 1$

Answer.

$\dfrac{(x-2)^2}{\dfrac{9}{4}} - \dfrac{(y+2)^2}{5} = 1$

Checkpoint A.9.39.

$(x+3)^2 - \dfrac{1}{5}(y+2)^2 = 1$

Answer.

$\dfrac{(x+3)^2}{3} - \dfrac{(y-2)^2}{\dfrac{3}{4}} = 1$

Checkpoint A.9.40.

$2x^2 - (y-1)^2 = \dfrac{8}{9}$

Answer.

$\dfrac{(x^2}{\dfrac{4}{9}} - \dfrac{(y-1)^2}{\dfrac{5}{9}} = 1$

Checkpoint A.9.41.

$\dfrac{2}{3}(x-5)^2 - \dfrac{5}{3}(y-6)^2 = \dfrac{6}{5}$

Answer.

$\dfrac{((x-5)^2}{\dfrac{9}{5}} - \dfrac{(y-6)^2}{\dfrac{18}{25}} = 1$

Subsubsection A.9.4.2 Find an asymptote

Example A.9.42.

The points $~(1,3),~(9,3),~(5,10),~$ and $~(5,-4)~$ are the midpoints of the four sides of a rectangle.

Sketch the rectangle.
Find equations for the diagonals of the rectangle.

Solution.

The easiest way find the diagonals of the rectangle is to sketch it first. From the sketch at right, we can see that the center of the rectangle is $(5,3)\text{,}$ and the two upper vertices are $(1,10)$ and $(9,10)\text{.}$
Both diagonals pass through the center, $(5,3)\text{.}$ The diagonal that passes through $(9,10)$ has slope
\begin{equation*} m=\dfrac{10-3}{9-5} = \dfrac{7}{4} \end{equation*}

$rectangle and diagoals$

We use the point-slope formula to find its equation.

\begin{align*} \dfrac{y-3}{x-5} \amp = \dfrac{7}{4} \amp \amp \blert{\text{Cross-multiply.}}\\ 4(y-3) \amp = 7(x-5) \amp \amp \blert{\text{Apply the distributive law.}}\\ 4y-12 \amp = 7x-35 \amp \amp \blert{\text{Add 12 to both sides.}}\\ 4y \amp = 7x-23 \amp \amp \blert{\text{Divide both sides by 4.}}\\ y \amp = \dfrac{7}{4}x - \dfrac{23}{4} \end{align*}

Similarly, you can check that the diagonal that passes through $(1,10)$ has slope $\dfrac{-7}{4}\text{,}$ and its equation is $y = \dfrac{-7}{4}x + \dfrac{47}{4}$

The four points given are the midpoints of the four sides of a rectangle.

Sketch the rectangle.
Find equations for the diagonals of the rectangle.

Checkpoint A.9.43.

$(-7,-1),~(1,-1),~(-3,-4),~(-3,2)$

Answer.

$rectangle and diagoals$
$\displaystyle y=\dfrac{3}{4}x+\dfrac{5}{4};~y=\dfrac{-3}{4}x-\dfrac{13}{4}$

Checkpoint A.9.44.

$(5,-8),~(9,-8),~(7,-15),~(7,-1)$

Answer.

$rectangle and diagoals$
$\displaystyle y=\dfrac{7}{2}x-\dfrac{65}{2};~y=\dfrac{-7}{2}x+\dfrac{61}{2}$

Subsection A.9.5 Nonlinear Systems

Subsubsection A.9.5.1 Write a system of equations

Problems that involve two variables can sometimes be described by a system of equations.

Example A.9.45.

The area of a rectangle is 874 square millimeters, and its perimeter is 122 millimeters. Write a system of equations for the dimensions of the rectangle.

Solution.

We use the formulas for area and perimeter, where $w$ stands for the width of the rectangle, and $l$ stands for its length.

\begin{align*} lw \amp = 874\\ 2l+2w \amp = 122 \end{align*}

Example A.9.46.

A small plane flies at a constant speed. In 5 hours, it travels 750 miles with a tailwind, but it travels only 600 miles in 5 hours against the wind. Write a system of equations for the speed of the plane and the speed of the wind.

Solution.

We let $v$ stand for the speed of the plane, and $w$ stand for the speed of the wind. We use the formula $RT=D\text{.}$

\begin{align*} (v+w)5 \amp = 750\\ (v-w)5 \amp = 600 \end{align*}

Write a system of equations for the problem.

Checkpoint A.9.47.

Darryl plans to mix some 50% solution with some 75% solution to make 20 liters of 60% solution. How much of each should he use?

Answer.

\begin{gather*} x+y=20\\ 0.5x+0.75y=12 \end{gather*}

Checkpoint A.9.48.

The perimeter of a rectangle is 38 inches. If we triple the width and decrease the length by 8 inches, we increase the area of the rectangle by 40%. What are the dimensions of the original rectangle?

Answer.

\begin{gather*} 2l+2w=38\\ 3w(l-8)=1.4lw \end{gather*}

Checkpoint A.9.49.

The express train travels 15 miles per hour faster than the local. The local takes 10 minutes longer to travel 30 miles than the express takes. Find the speed of each.

Answer.

\begin{gather*} v=15+l\\ \dfrac{30}{l} = \dfrac{30}{v}+\dfrac{1}{6} \end{gather*}

Checkpoint A.9.50.

Delbert made $65 interest on his money market account this year. Francine invested $100 less than Delbert at 0.5% higher interest rate and earned $66. How much did Delbert invest, and at what interest rate?

Answer.

\begin{gather*} Pr=65\\ (P-100)(r+0.005) = 66 \end{gather*}

Subsubsection A.9.5.2 Use substitution

We can use substitution to solve some systems of equations.

Example A.9.51.

Use substitution to write an equation in one variable.

\begin{align*} x^2-8x+y^2+2y \amp = 23\\ x+2y \amp = 12 \end{align*}

Solution.

We solve the second equation for $x$ to get $x = \alert{12-2y}\text{,}$ and then substitute this expression into the first equation.

\begin{align*} (\alert{12-2y})^2-8(\alert{12-2y})+y^2+2y \amp = 23\\ 144-48y+4y^2-96+16y+y^2+2y \amp = 23\\ 5y^2-30y+25 \amp = 0 \end{align*}

Use substitution to write an equation in one variable.

Checkpoint A.9.52.

\begin{align*} x^2-2x+y^2+4y \amp = 20\\ y-x \amp = 4 \end{align*}

Answer.

$2x^2+10x+12=0$

Checkpoint A.9.53.

\begin{align*} x^2-y^2 \amp = 16\\ xy \amp = 6 \end{align*}

Answer.

$x^4-16x^2-36=0$

Checkpoint A.9.54.

\begin{align*} x^2-12x+y^2+2 \amp = 0\\ y \amp = 4x-7 \end{align*}

Answer.

$17x^2-68x+51=0$

Checkpoint A.9.55.

\begin{align*} x^2-4x+y^2+2y \amp = 5\\ x+2y \amp = -5 \end{align*}

Answer.

$5y^2+30y+40=0$

Example A.9.56.

Find the intersection points of the graphs of

\begin{align*} x^2+y^2\amp = 5\\ xy \amp = 2 \end{align*}

Solution.

We will use substitution to solve the system. First we solve the "easier" of the two equations (the second equation) for $y$ to obtain

\begin{equation*} y=\dfrac{2}{x} \end{equation*}

We substitute $\alert{\dfrac{2}{x}}$ for $y$ in the first equation to find

\begin{equation*} x^2 + (\alert{\dfrac{2}{x}})^2 = 5 ~~~~\text{or} ~~~~~ x^2 + \dfrac{4}{x^2}=5 \end{equation*}

This equation has only one variable, $x\text{,}$ and we solve it by first clearing fractions. We multiply both sides by $x^2\text{,}$ and then subtract $5x^2$ to obtain

\begin{equation*} x^4-5x^2+4=0 \end{equation*}

Then we factor the left side to get

\begin{equation*} (x^2-1)(x^2-4)=0 \end{equation*}

and apply the zero-factor principle to find

\begin{equation*} x^2-1 = 0 ~~~~ \text{or} ~~~~ x^2-4=0 \end{equation*}

We solve each of these equations to find

\begin{equation*} x=1,~~ x=-1, ~~ x=2, ~~ \text{or} ~~ x=-2 \end{equation*}

Finally, we substitute each of these values into $y=\dfrac{2}{x}$ to find the $y$-components of each solution. The intersubsection points of the two graphs are $(1,2),~ (-1,-2), ~ (2,1)$ and $(-2,-1)\text{.}$ The graph of the system is shown below.

Checkpoint A.9.57.

Find the intersection points of the graphs of

\begin{align*} x^2-y^2\amp = 35\\ xy \amp = 6 \end{align*}

Answer.

$(6,1),~ (-6,-1)$

Subsubsection A.9.5.3 Use elimination

Example A.9.58.

Find the solutions to the following system of the equations.

\begin{align*} x^2-2y^2\amp = 1\\ \dfrac{x^2}{15}+\dfrac{y^2}{10} \amp = 1 \end{align*}

Verify the solutions on a graph.

Solution.

We multiply the first equation by 3 and the second by 60 to obtain the new system

\begin{align*} 3x^2-6y^2\amp = 3\\ 4x^2+6y^2 \amp = 60 \end{align*}

Adding these two equations, we have

\begin{align*} 7x^2\amp = 63\\ x^2 \amp = 9\\ x \amp = \pm 3 \end{align*}

We substitute these values for into any of the equations involving $y$ and solve, to find the solutions $(3,2), ~ (3,-2),~ (-3,2)$ and $(-3,-2)\text{.}$

The two original equations describe a hyperbola and an ellipse. We can obtain graphs on a graphing utility by solving each equation for $y$ to get

\begin{align*} y\amp = \pm\sqrt{\dfrac{x^2-1}{2}}\\ y \amp = \pm\sqrt{10(1-\dfrac{x^2}{15}} \end{align*}

We enter these equations to obtain the graph shown below. The solutions of the system are the intersubsection points of the graphs.

Checkpoint A.9.59.

Find the intersection points of the graphs of

\begin{align*} x^2-y^2\amp = 5\\ x^2 + y^2 \amp = 13 \end{align*}

Answer.

$(2,3),~ (-2,3),~ (2,-3),~ (-2,-3)$

For some quadratic systems, we can use a combination of elimination of variables and substitution.

Example A.9.60.

Find the intersection of the circles given by the equations.

\begin{align*} x^2 - 4x + y^2 + 2y\amp = 20\\ x^2 - 12x + y^2 + 10y \amp = -12 \end{align*}

Solution.

We subtract the second equation from the first to obtain

\begin{equation*} 8x-8y=32 \end{equation*}

Solving for $x$ we have

\begin{equation*} x=y+4 \end{equation*}

We substitute $\alert{y+4}$ for $x$ into either of the original equations. We’ll use the first equation.

\begin{align*} (\alert{y+4})^2 - 4(\alert{y+4}) + y^2 + 2y\amp = 20\\ (y^2+8y+16)- 4y-16 +y^2 + 2y \amp = 20\\ 2y^2+6y-20 \amp = 0\\ y^2+3y-10 \amp = 0\\ (y+5)(y-2) \amp = 0 \end{align*}

Thus $y=-5$ or $y=2\text{.}$ From our first new equation we find that when $y=-5,~ x=-1\text{,}$ and when $y=2,~ x=6\text{.}$ Thus the two circles intersect at $(-1,-5)$ and $(6,2)\text{,}$ as shown below.

Checkpoint A.9.61.

Find the intersection points of the graphs of

\begin{align*} x^2-8x + y^2 + 2y\amp = 23\\ x^2 -12x + y^2 - 6y \amp = -25 \end{align*}

Answer.

$(2,5),~ (10,1)$

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