Apples sell for $1.90 per pound and bananas sell for $0.75 per pound. Troy bought some apples and some bananas. Together they weighed 3.8 pounds and cost $5.84. How many pounds of apples and how many pounds of bananas did Troy buy?
1.9 pounds of apples; 1.9 pounds of bananas

1.2 pounds of apples; 2.6 pound of bananas

1.5 pounds of apples; 2.3 pounds of bananas

2.6 pounds of apples; 1.2 pounds of bananas
Raisins sell for $3.50 per pound and granola sells for $5.90 per pound. Terri bought some raisins and some granola. The total weight was 2.1 pounds and the cost was $8.79.

How many pounds of raisins and how many pounds of granola did Terri buy?

0.6 pounds of raisins; 1.5 pounds of granola

1 pound of raisins; 1.1 pounds of granola

1.5 pounds of raisins; 0.6 pounds of granola

1.1 pounds of raisins; 1 pound of granola
QUESTION 3

Almonds sell for $4.50 per pound and green beans for $3.75 per pound. Paul bought some almonds and some green beans. The total weight was 3.5 pounds and the cost was $9.75.

Let p represent the number of pounds of almonds.

Which equation represents the situation described?

3.75p + 9.75(3.5 – p) = 4.5

3.75p + 4.5(p – 3.5) = 9.75

4.5p + 3.75(3.5 – p) = 9.75

4.5p + 3.75(p – 3.5) = 9.75
QUESTION 4

Alex works as an office clerk and his office job pays him $11 per hour. Monica works as a cashier and her job pays her $12.75 per hour. Together, they worked 45 hours and earned a total of $530.

How many hours did Monica work in her job?

30

20

25

35
Soren has two jobs. During the day, he works as an office clerk. In the evening, he works as a cashier. His office job pays him $11 per hour. His cashier job pays him $8.75 per hour. In one week, Soren worked 46 hours. He earned a total of $470.

Let c represent the number of hours that Soren works as an office clerk. Which equation can be used to find how many hours Soren worked in both jobs?

8.75c + 11(46 – c) = 470

11c + 8.75(470 – c) = 46

11c + 8.75(46 – c) = 470

11c + 8.75(c – 46) = 470
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ummmmmmmmmmm how many questions is this ummmmmmmmmmmmm ok .the first question answer is 1.5 the third one and the second question is the last one and the third question is........................................ 20 and the last answer to the last question isssssssssssssssss...................... the last one to. ok yourwelcome have a nice day. :]

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martinleallende martinleallende · Answer 2 · 2019-09-10T22:36:53+02:00

Answer:

Q1 =2.6 pounds of apples, 1.2 pounds of banana.

Q2=1.5 pounds of raisins, 0.6 pounds of granola.

Q3=4.5p+3.75(3.5-p)=9.75.

Q4=20.

Q5=11c+8.75(46-c)=470

Step-by-step explanation:

Let's begin the explanation. These problems are solved by using reduction, equalization or substitution method needed to solve linear system of two equations.

Q1. You have to state the facts: $1.90 per pound of apples, $0.75 per pound of bananas, total weight 3.8 pounds, total cost $5.84.

Next, let's name things, beign pounds of apples = x and pounds of bananas = y.

Now, you have to form the linear system with what you know, and what you've named:

[tex]\left \{ {{1.9x+0.75y=5.84} \atop {x+y=3.8}} \right.[/tex]

Now you solve. Here I'm going to explain the equalization method.

First thing you need to do is to solve one of the unknowns in both equations. (It has to be the same unknown.)

In this case we will go on solving X from both equations.

[tex]Equation1 : 1.9x+0.75y=5.84\\ 1.9x=5.84-0.75y\\x=(5.84-0.75y)/1.9\\x=3.11-0.39y\\\\Equation 2 : x+y=3.8\\x=3.8-y[/tex]

Next you have to equalizate the expressions obtained, and solve them:

[tex]3.11-0.39y=3.8-y\\-0.39y+y=3.8-3.11\\0.61y=0.69\\y=0.69/0.61\\y≅1.2[/tex] It's not exactly 2 because I've rounded decimals. But if you do it straight forward it will give you 1.2.

So, now you have the value of bananas pounds (1.2). Next step is to substitute it in any of the equations of the system, and you will get the value of apples pounds.

[tex]x+y=3.8\\x+1.2=3.8\\x=3.8-1.2\\x=2.6[/tex]

The solution to question 1 is 2.6 pounds of apples; 1.2 pounds of bananas.

Q2: Similarly to what you have done before, this time x=pounds of raisins, y=pounds of granola.

[tex]\left \{ {{3.5x+5.9y=8.79} \atop {x+y=2.1}} \right.[/tex]

We will solve this by the substitution method.

Step 1 - Solve 1 equation:

[tex]x+y=2.1\\x=2.1-y[/tex]

Step 2 - Replace it in the other equation:

[tex]3.5x+5.9y=8.79\\3.5(2.1-y)+5.9y=8.79[/tex]

Step 3 - Solve the equation:

[tex]3.5(2.1-y)+5.9y=8.79\\7.35-3.5y+5.9y=8.79\\-3.5y+5.9y=8.79-7.35\\2.4y=1.44\\y=1.44/2.4\\y=0.6[/tex]

Step 4 - Replace the value of y in any equation:

[tex]x+y=2.1\\x+0.6=2.1\\x=2.1-0.6\\x=1.5[/tex]

The solution to question 2 is 1.5 pounds of raisins; 0.6 pounds of granola.

Q3 Similarly to what you have done before, this time p=pounds of almonds, g=pounds of green beans.

This one is solved by the substitution method explained before.

We set the linear system

[tex]\left \{ {{4.5p+3.75g=9.75} \atop {p+g=3.5}} \right.[/tex]

Step 1 - Solve 1 equation:

[tex]p+g=3.5\\g=3.5-p[/tex]

Step 2 - Replace it in the other equation:

[tex]4.5p+3.75g=9.75\\4.5p+3.75(3.5-p)=9.75[/tex]

You don't need to go on solving this one.

The solution to question 3 is 4.5p+3.75(3.5-p)=9.75

Q4 Once again, state the facts, and set the linear system:

x=Alex work hours, y=Monica work hours

[tex]\left \{ {{11x+12.75y=530} \atop {x+y=45}} \right.[/tex]

We will solve this by using the reduction method

Step 1 - Make one pair of coefficients of the same variable in negatives of one another.

[tex]\left \{ {{11x+12.75y=530} \atop -11*({x+y=45)}} \right. \\\\\left \{ {{11x+12.75y=530} \atop {-11x-11y=-495}} \right.[/tex]

Step 2 - Add the equations

[tex]\left \{ {{11x+12.75y=530} \atop {-11x-11y=-495}} \right.\\ 1.75y=35[/tex]

Step 3 - Solve the new equation

[tex]1.75y=35\\y=35/1.75\\y=20[/tex]

You could continue on, replacing Y value in any equation to get X value, but for this exercise we're done.

The solution to question 4 is 20 hours.

Q5 As we have done before, state the facts, set the linear system:

c=hours worked as an office clerk y=hours worked as a cashier

[tex]\left \{ {{11c+8.75y=470} \atop {c+y=46}} \right.[/tex]

Remember this kind of exercises are solved by using the substitution method

Step 1 - Solve 1 equation:

[tex]c+y=46\\y=46-c[/tex]

Step 2 - Replace it in the other equation:

[tex]11c+8.75y=470\\11c+8.75(46-c)=470[/tex]

You don't need to go on solving this one, but in case you want to

Step 3 - Solve the equation:

[tex]11c+8.75(46-c)=470\\11c+402,5-8.75c=470\\11c-8.75c=470-402.5\\2.25c=67.5\\c=67.5/2.25\\c=30[/tex]

Step 4 - Replace the value in any equation:

[tex]c+y=46\\30+y=46\\y=40-30\\y=16[/tex]

The solution to question 5 is 11c+8.75(46-c)=470 and he has worked 30 hours as an office clerk and 16 as a cashier

I hope I've helped you, have a nice day.