Kx^2+k=8x-2xk

Mathematics

1 answer:

satela [25.4K]3 years ago

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Answer:

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A business has two loans totaling $50,000. One loan has a rate of 8% and the other has a rate of 12%. This year, the business ex

Debora [2.8K]

Answer:

the 8% loan has a principal of $37500

the 12% loan has a principal of $12500

Step-by-step explanation:

Let's start by writing the general equation for the interest hwre I is the interest, P is the principal (in our case would be loan amounts), "r" is the interest rate in decimal form (in our case one would be 0.12, and the other one 0.08), and t is the time in years (in our case 1 year).

$I=P*r*t$

Then we write the interest equation coming from each loan at the end of this year (we call I1 the interest coming from the 12% loan and I2 the interest coming from the 8% one). Since we don't know the loan amounts (in fact those are what we need to find) we will name one "x" and the other "y":

$I=P*r*t\\I1=x * 0.12*1\\I2=y*0.08*1$

Next, we add these last two equations term by term, and replace the addition of both interests by $4500 as given in the information:

$I1=x * 0.12*1\\I2=y*0.08*1\\I1+I2 = 0.12x+0.08y\\4500=0.12x+0.08y$

This is our first equation in the variables x and y which are our unknowns.

Now we generate the second equation on x and y by writing in agebraic terms the other piece of information we have: "the total of the two loans is $50000. That is the addition of the principals x and y should equal $50000:

$x+y=50000$

We solve for y in this last equation and replace its form in terms of x in the equation of the interest, and solve for the unknown x:

$y=50000-x\\4500 = 0.12x +0.08 y\\4500=0.12x+0.08(50000-x)\\4500=0.12x+4000-0.08x\\4500=0.12x-0.08x+4000\\4500=0.04x+4000\\4500-4000=0.04x\\500=0.04x\\x=\frac{500}{0.04} =12500$