Question

problem 9:
a system of two equations has the solution (6, 2).
here is a graph of the one of the equations:
what could the other equation be?
a. ( y = 4x - 2 )
b. ( y = \frac{1}{2}x - 1 )
c. ( y = \frac{2}{3}x - 1 )

problem 10a:
ari and kiri are each saving money.

• ari starts with $100 in savings, and saves $5 each week
• kiri starts with $40 in savings and saves $10 each week

after 4 weeks, who has more money in savings?

problem 10b:
after how many weeks will ari and kiri
have the same amount of money in
savings?
solve through equations or using the
provided graphing space.
( 5x + 100 = 10x + 40 )

Explanation:

Problem 9

Step1: Test point (6,2) in option a

Substitute $x=6, y=2$ into $y=4x-2$:
$2 = 4(6)-2 = 24-2=22$ → $2≠22$, invalid.

Step2: Test point (6,2) in option b

Substitute $x=6, y=2$ into $y=\frac{1}{2}x-1$:
$2 = \frac{1}{2}(6)-1 = 3-1=2$ → $2=2$, valid.

Step3: Test point (6,2) in option c

Substitute $x=6, y=2$ into $y=\frac{2}{3}x-1$:
$2 = \frac{2}{3}(6)-1 = 4-1=3$ → $2≠3$, invalid.

Problem 10a

Step1: Calculate Ari's savings after 4 weeks

Ari's savings: $100 + 5\times4$
$=100+20=120$

Step2: Calculate Kiri's savings after 4 weeks

Kiri's savings: $40 + 10\times4$
$=40+40=80$

Step3: Compare the two amounts

$120>80$, so Ari has more.

Problem 10b

Step1: Rearrange the given equation

Start with $5x + 100 = 10x + 40$
Subtract $5x$ from both sides: $100 = 5x + 40$

Step2: Isolate the term with x

Subtract 40 from both sides: $100-40=5x$
$60=5x$

Step3: Solve for x

Divide both sides by 5: $x=\frac{60}{5}=12$

Answer:

Problem 9: b. $y = \frac{1}{2}x - 1$
Problem 10a: Ari has more money in savings (Ari has $120, Kiri has $80)
Problem 10b: 12 weeks