Andre is in charge of cooking broccoli and zucchini for a large group. He has to spend all $17 he has and can carry 10 pounds of veggies. Zucchini costs $1.50 per pound and broccoli costs $2 per pound. One graph shows combinations of zucchini and broccoli that weigh 10 pounds and the other shows combinations of zucchini and broccoli that cost $17. a. Name one combination of veggies that weighs 10 pounds but does not cost $17.

b. Name one combination of veggies that costs $17 but does not weigh 10 pounds.

c. How many pounds each of zucchini and broccoli can Andre get so that he spends all $17 and gets 10 pounds of veggies?

Andre has 4 pounds of Broccoli and 6 pounds of Zucchini.

Step-by-step explanation:

In this question there is mistake in given data:

With given data of $117 and 10 pounds of veggies solution is not possible

To make possible solution we should make either data $17 or we should make 100 pounds.

We have solved by making $17.

Given:

Amount of veggies he can carry = 10 pounds

Let the amount of broccoli be 'x'.

Let the amount of Zucchini be 'y'.

Now we can say that;

Amount of veggies he can carry is equal to sum of the amount of broccoli and the amount of Zucchini.

framing in equation form we get;

x+y=10 \ \ \ \ equation \ 1x+y=10 equation 1

Also Given:

Total money he has = $17

Per pound of Zucchini costs = $1.50

Per pound of broccoli cost = $2

Now we can say that;

Total money he has is equal to sum of the amount of broccoli multiplied by Per pound of broccoli cost and the amount of Zucchini multiplied by Per pound of Zucchini costs.

framing in equation form we get;

2x+1.5y =17 \ \ \ \ equation \ 22x+1.5y=17 equation 2

Now Multiplying equation 1 by 1.5 we get;

\begin{gathered}1.5(x+y)=10\times1.51.5x+1.5y =15 \ \ \ \ equation \ 3\end{gathered}

1.5(x+y)=10×1.5

1.5x+1.5y=15 equation 3

Subtracting equation 3 from equation 2 we get;

\begin{gathered}2x+1.5y-(1.5x+1.5y)=17-152x+1.5y-1.5x-1.5y=20.5x=2\end{gathered}

2x+1.5y−(1.5x+1.5y)=17−15

2x+1.5y−1.5x−1.5y=2

0.5x=2

Dividing both side by 0.5 we get;

\begin{gathered}\frac{0.5x}{0.5}=\frac{2}{0.5}x=4\ pounds\end{gathered}

0.5

0.5x

=

0.5

2

x=4 pounds

Now substituting the value of 'x' in equation 1 we get;

\begin{gathered}x+y=104+y=10y=10-4 = 6\ pounds\end{gathered}

x+y=10

4+y=10

y=10−4=6 pounds

Hence Andre has 4 pounds of Broccoli and 6 pounds of Zucchini.

0.02040816326