Math  /  Algebra

Question22. Michelle Is making goodie bags for Christmas filled with chocolates and hard candies. Chocolates cost How per lb , and hard candies cost $3.00\$ 3.00 per 1 lb . Michelle bought a total of 15 lbs . and spent a total of $40\$ 40. How many lbs, of each type did Michelle purchase? b) Define your variables. b) Write an equation showing that Michelle spent a total of $40\$ 40 on chocolates and hard candies. c) Write an equation showing that Michelle bought a total of 15 lbs . of chocolates and hard candy. d) Solve the system of equations to find out how many lbs. of chocolates and how many lbs. of hard candies Michelle bought. c=2.50151bsh=3.00=40$\begin{aligned} c & =2.50 \quad 151 b s \\ h= & 3.00 \\ & =40^{\$} \end{aligned}
Chocolates == \qquad Hard Candies = \qquad
23. Your teacher is giving a test worth 100 points. The test contains a total of 40 questions. Some questions are worth 2 points and some questions are worth 4 points on the test. How many of each type of question are on the est?

100 409 2 point problems == \qquad 4 point problems = \qquad

Studdy Solution
Solve the system of equations to find out how many pounds of chocolates and how many pounds of hard candies Michelle bought.
We have the system of equations:
1. 2.50x+3.00y=40 2.50x + 3.00y = 40
2. x+y=15 x + y = 15

First, solve the second equation for x x :
x=15y x = 15 - y
Substitute x=15y x = 15 - y into the first equation:
2.50(15y)+3.00y=40 2.50(15 - y) + 3.00y = 40
Expand and simplify:
37.52.50y+3.00y=40 37.5 - 2.50y + 3.00y = 40
Combine like terms:
37.5+0.50y=40 37.5 + 0.50y = 40
Subtract 37.5 from both sides:
0.50y=2.5 0.50y = 2.5
Divide both sides by 0.50:
y=2.50.50 y = \frac{2.5}{0.50} y=5 y = 5
Now, substitute y=5 y = 5 back into x=15y x = 15 - y :
x=155 x = 15 - 5 x=10 x = 10
Michelle bought:
10 pounds of chocolates and 5 pounds of hard candies.

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