Question

3.2.2 hw

1. what is the area of each part of the entire rectangle at right? what is the area of the whole figure? write the area of the rectangle as a product and as a sum.

x -5
x 7
x² - 12x + 2=

1. solve for the variable in each equation below. show your work and check your solutions.

a. -4x + 7 = 31
b. 7b - 11 = 10b - 41
c. 3/4x = 3/2
d. 9a - 15 = 10a + 7

1. write and solve an equation to represent the given situation. be sure to define your variable.

samantha currently has $1600 in the bank and is spending $95 per week. how many weeks will it take until her account is worth only $1220?

1. graph the line y = 2/5x on graph paper.

a. draw a slope triangle.
b. rotate your slope triangle 90° around the origin to get a new slope triangle. what is the new slope?
c. write the equation of a line perpendicular to y = 2/5x

Explanation:

Step1: Solve part 1 - area of rectangle parts and whole

The rectangle is divided into four parts. The areas of the four parts are: $x\times x=x^{2}$, $x\times7 = 7x$, $- 5\times x=-5x$, $-5\times7=-35$.
The area of the whole rectangle as a product is $(x - 5)(x + 7)$. Using the FOIL method: $(x-5)(x + 7)=x\times x+x\times7-5\times x - 5\times7=x^{2}+7x-5x - 35=x^{2}+2x-35$. As a sum, it is $x^{2}+7x-5x - 35$.

Step2: Solve part 2 - solve equations

• a. Solve $-4x + 7=31$

Subtract 7 from both sides: $-4x+7 - 7=31 - 7$, so $-4x=24$. Divide both sides by - 4: $x=\frac{24}{-4}=-6$.

• b. Solve $7b-11 = 10b-41$

Subtract 7b from both sides: $7b-11-7b=10b - 41-7b$, so $-11 = 3b-41$. Add 41 to both sides: $-11 + 41=3b-41 + 41$, so $30 = 3b$. Divide both sides by 3: $b = 10$.

• c. Solve $\frac{3}{4}x=\frac{3}{2}$

Multiply both sides by $\frac{4}{3}$: $x=\frac{3}{2}\times\frac{4}{3}=2$.

• d. Solve $9a-15=10a + 7$

Subtract 9a from both sides: $9a-15-9a=10a + 7-9a$, so $-15=a + 7$. Subtract 7 from both sides: $a=-15 - 7=-22$.

Step3: Solve part 3 - word - problem

Let $w$ be the number of weeks. Samantha starts with $1600$ and spends $95$ per week. The equation is $1600-95w=1220$.
Subtract 1600 from both sides: $1600-95w-1600=1220 - 1600$, so $-95w=-380$. Divide both sides by - 95: $w=\frac{-380}{-95}=4$.

Step4: Solve part 4 - graphing and slope problems

• **a. For $y=\frac{2}{5}x$, choose two points. Let $x = 0$, then $y = 0$; let $x = 5$, then $y = 2$. Draw a line through $(0,0)$ and $(5,2)$ and draw a slope - triangle with vertical side 2 and horizontal side 5.
• **b. When a line with slope $m$ is rotated 90° around the origin, the new slope $m'$ is the negative reciprocal of the original slope. The original slope $m=\frac{2}{5}$, so the new slope $m'=-\frac{5}{2}$.
• **c. The slope of a line perpendicular to $y=\frac{2}{5}x$ (with slope $\frac{2}{5}$) is the negative reciprocal. The equation of a line in slope - intercept form is $y=mx + b$. The slope of the perpendicular line is $-\frac{5}{2}$. If it passes through the origin (assuming no other point given), the equation is $y=-\frac{5}{2}x$.

Answer:

1. Areas of parts: $x^{2},7x,-5x,-35$; Area of whole rectangle as product: $(x - 5)(x + 7)$; as sum: $x^{2}+2x-35$
2. a. $x=-6$; b. $b = 10$; c. $x = 2$; d. $a=-22$
3. $w = 4$ weeks
4. a. Draw line through $(0,0)$ and $(5,2)$ and slope - triangle; b. New slope is $-\frac{5}{2}$; c. $y=-\frac{5}{2}x$