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A roadside fruit stand sells apples at 75 cents a pound, peaches at 90 cents a pound, and pears at 60 cents a pound. Muriel buys 18 pounds of fruit at total cost of 13.80 dollars. Her peaches and pears together cost 1.80 more than her apples. Set up a linear system for the number of pounds of apples, peaches and pears that she bought.

I have made an equation and I'm not so sure about it. I know that there is something's wrong with this equation. Can you tell me what is it? Here it is: $$ 75x+90y+60z=1380 $$ $$ x+y+z=18 $$ $$ -1.80x+y+z=0 $$

kjetil b halvorsen
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user114811
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2 Answers2

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Your last equation is wrong.

As you identified in your first equation, the price of apples is $75x$, the price of peaches $90y$ and the price of pears $60z$.

Her peaches and pears together cost 1.80 more than her apples

The price of apples plus $180$ cents is the same as the price of peaches and pears together.

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Let us assume that Muriel has bought $x$ pounds of apples, $y$ pounds of peaches and $z$ pounds of pears. Assuming $1$ pound = $100$ cents (I do not know the exact conversion),

$$x+y+z = 18$$ $$0.75x + 0.90y + 0.60z = 13.80$$ $$0.75x + 1.80 = 0.90y + 0.60z$$

NOTE: It is mentioned that "peaches and pears together cost $1.80$ more than her apples". Also it should have been $90y+60z$ in your last equation as we are talking about the prices.

lsp
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