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3 years ago

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Solve the system by substitution 2.5x-3y=-13 3.25x-y=-14

2 answers:

Degger [83]3 years ago

4 0

Solve for the first variable in one equation then substitute the results into the other equation. (-4,1)

taurus [48]3 years ago

3 0

Answer:

$(-4,1)$

Step-by-step explanation:

we have

$2.5x-3y=-13$ -----> equation A

$3.25x-y=-14$ -----> isolate variable y

$y=3.25x+14$ ------> equation B

Substitute equation B in equation A

$2.5x-3[3.25x+14]=-13$

$2.5x-9.75x-42=-13$

$-7.25x=-13+42$

$-7.25x=29$

$x=-4$

Find the value of y

$y=3.25(-4)+14=1$

the solution is the point $(-4,1)$

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Answer:

<em>a. The system of equations that models the situation is....</em>

<em> $0.80x+0.50y=3.60\\\\ 0.20x+0.50y=2.40$ </em>

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Step-by-step explanation:

Suppose, the amount of 80/20 mixture is $x$ pounds and the amount of 50/50 mixture is $y$ pounds.

So, the amount of peanuts in 80/20 mixture $= 0.80x$ pound and the amount of almonds in 80/20 mixture $=0.20x$ pound.

And the amount of peanuts in 50/50 mixture $=0.50y$ pound and the amount of almonds in 50/50 mixture $=0.50y$ pound.

Now, Sarah would like to make a 6 pounds nut mixture that is 60% peanuts and 40% almonds.

So, the amount of peanuts in that mixture $=(6\times 0.60)=3.60$ pounds

and the amount of almonds in that mixture $=(6 \times 0.40)= 2.40$ pounds.

So, the system of equations will be.........

$0.80x+0.50y=3.60 ...................(1)\\\\ 0.20x+0.50y=2.40...................(2)$

Subtracting equation (2) from equation (1), we will get.....

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Now, plugging this $x=2$ into equation (1), we will get......

$0.80(2)+0.50y=3.60\\ \\ 1.60+0.50y=3.60\\ \\ 0.50y=3.60-1.60=2\\ \\ y=\frac{2}{0.50}=4$

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