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Homework 1
Section 2.2 #6: Solve

$7x+2=3x-9$ .
Solution: Subtract

$3x$ to get

$4x+2=-9$ . Subtract

$2$ to get

$4x=-11$ . Divide by

$4$ to get

$x = -\dfrac{11}{4}$ .

Section 2.2 #7: Solve

$4x-3=5$ .
Solution: Add

$3$ to get

$4x=8$ . Divide by

$4$ to get

$x=2$ .

Section 2.2 #8: Solve

$3(x+2)-12=5(x+1)$ .
Solution: Distribute the

$3$ on the left and the

$5$ on the right to get

$3x+6-12=5x+5$ . Simplify the arithmetic on the left to get

$3x-6=5x+5$ . Add

$6$ to get

$3x=5x+11$ . Subtract

$5x$ to get

$-2x=11$ . Divide by

$-2$ to get

$x=-\dfrac{11}{2}$ .

Section 2.5 #8: Solve by factoring:

$2x^2+9x-5=0$ .
Solution: The left-hand side factors as

$(2x-1)(x+5)=0.$ This means

$2x-1=0$ or

$x+5=0$ . Solving each of these yields

$x=\dfrac{1}{2}$ and

$x=-5$ .

Section 2.5 #19: Solve the equation

$x^2=36$ .
Solution: Taking square roots of each side yields

$x=\pm \sqrt{36} = \pm 6$ .

Section 2.5 #38: Solve

$2x^2+5x+3=0$ .
Solution: It is not clear how to factor this. Therefore apply the quadratic formula to get

$x=\dfrac{-5 \pm \sqrt{25-4(2)(3)}}{2(2)} = \dfrac{-5 \pm \sqrt{1}}{4} = \dfrac{-5 \pm 1}{4}.$ Therefore the solutions are

$x=\dfrac{-4}{4}=-1$ and

$x=\dfrac{-6}{4}=-\dfrac{3}{2}$ .

Section 3.5 #29: Sketch the graph of

$k(x)=(x-2)^3-1$
Solution: Notice that there is a horizontal shift to the right by

$2$ and a vertical shift down by

$1$ . Draw it in the following way:

Section 7.1 #8: Draw an angle in standard position with the measure

$-80^{\circ}$ .
Solution:

Section 7.1 #11: Draw an angle in standard position with the measure

$\dfrac{2\pi}{3}$ .
Solution: