two real numbers are such that the sum of 3 times one plus 4 times the other is 68, while the product of the numbers is 80. find the numbers

Question

i dont really get how to do this problem.

Answer 1

Hi Jarnail,

 

This problem starts with the fact that you have 2 variables and therefore need two equations.

 

Equation 1:

3*x+4*y=68   (Where x is one of your real numbers and y is your second real number)

 

Equation 2:

x*y=80           (Product of your numbers is 80...)

 

Now, we need to do substitution... I chose to solve Equation 2 for y.

 

x*y=80

x*y/x = 80/x   (divide both sides by variable x, to get y by itself)

 

y=80/x          (Equation 2a)

 

Now substitute Equation 2a into Equation 1. 

 

3*x+4*(80/x) = 68    (Equation 1a)

 

We would get from distributing,

 

3*x+320/x = 68     (We do not like have variables in the denominator, so multiple the entire equation by 'x')

 

3*x² + 320 = 68*x  (Set the equation equal to 0 so that we can factor)

-68*x            -68*x

------------------------

3x²-68x+320 = 0     (now we can factor, my preferred method because it always works is the quadratic formula)

 

For ax²+bx+c = 0

 

x = (-b ± √(b²-4*a*c))/(2*a)          (Quadratic Formula)

 

For us, a=3, b=-68, and c = 320. We now need to plug and chug into the equation. 

 

x = (--68±√((-68)² - 4*3*320))/(2*3)     (68²=4624, 4*3*320 = 3840)

 

x = (68 ±√(4624 -3840))/ 6

 

x = (68±√(784))/6                                (√784 =28)

 

x = (68±28)/6 

 

x = (68-28)/6  and  x = (68+28)/6

 

x = 6.666667  and  x = 16       (We can use either of these solutions, however, I would rather not do fractions so I                                                   picked x=16)

 

With x = 16 we now use Equation 2a to solve for y

 

y = 80/x

 

y = 80/ 16

 

y = 5

 

so we now have a solution, x = 16 and y = 5

 

To make sure that our solution works, we need to check it with Equation 1.

 

3*16+4*5 = 68 ?

 

48+20=68

 

68=68     (Yay!!!) 

 

Our solution is correct!

 

I hope this helped, let me know if you have any questions

Answer 2

Who knows what the two real numbers are? No one knows yet, so for now, just call them "a" and "b." 3 times of the numbers plus 4 times the other is 68. Let's say "3 times a" and "b times 4". The addition of those products will get a sum of 68.

 

3a + 4b = 68

 

Since the product of the numbers is 80, that means a times b is 80

 

ab = 80

 

We made two equations. See how nice they look next to each other.

 

3a + 4b = 68

ab = 80

 

This is called solving for a system of equations. One of the ways to do this is to use substitution. Check this out.

 

ab = 80

b = 80/a

 

Because b = 80/a, substitute this in for b in the addition equation.

 

3a + 4b = 68

3a + 4(80/a) = 68

 

Now this whole equation is in terms of just one variable, so that will make solving easier. Get a alone to solve for it.

 

3a + 4(80/a) = 68

3a + 320/a = 68

 

The second term is a fraction. Like when it comes to adding any other fraction, common denominators is the way to go. The common denominator is a, so 3a is going to need a denominator of a, so it will look like 3a²/a

 

3a + 320/a = 68

3a²/a + 320/a = 68

 

Combine the left side into one big fraction like this.

 

(3a² + 320)/a = 68

 

From here, notice there's an a² term. That means there's probably going to be a quadratic equation that comes out of this. It's easiest to solve those when it's in standard form, which looks like ax² + bx + c = 0. To get this in standard form, multiply both sides by a to make the fraction not a fraction anymore.

 

(3a² + 320)/a = 68

3a² + 320 = 68a

 

Again, when solving quadratic equations, one side must be equal to zero, so subtract 68a from both sides.

 

3a² - 68a + 320 = 0

 

The coefficient of the squared term is 3. To factor quadratics when that coefficient is not 1, multiply this with the c term, so multiply (3)(320), which is 960. This means we need to find two numbers that multiply to 960 and add to -68. Those numbers are -48 and -20. Call these our magic numbers. Now write this out.

 

(3a - 48)(3a - 20) = 0

 

Those terms of 3a are there because the original equation had a coefficient of 3 in front of the squared term. The -48 and -20 are the two magic numbers.

 

Look at (3a - 48). If there's a GCF between 3x and 48, divide both terms by that GCF. In this case, that GCF is 3, so divide both of those terms by 3.

 

(3a - 48)(3a - 20) = 0

(a - 16)(3a - 20) = 0

 

Based on this equation x = 16 and 20/3. Most likely, your teacher was probably asking for the positive whole number, which is 16, but I'll work with both.

 

Plug a = 16 into one of the equations way back in the beginning. In this case, I'll use this equation.

 

ab = 80

16b = 80

b = 5

 

This means that if a = 16, then b = 5. One of the answers for a was also 20/3. Do the same thing we did.

 

ab = 80

(20/3)b = 80

b = 12

 

All in all, if a = 16 then b = 5. If a = 20/3, then b = 12. I hope this helps, and good luck with your math class.

 

Answer 3

3x + 4y = 68

x * y  =80

 

# Isolate for x.

x = 80/y

 

# Now plug into the other formula.

3(80/y) + 4y = 68

240/y + 4y = 68

240 + 4y^2 = 68y

4y^2 - 68y + 240 = 0

y^2 - 17y + 60 = 0

(y-12)(y-5) = 0

y = 12, 5

 

#now plug these in to solve for x

x = 80/12 = 6.66

 

x = 80/5 = 16

 

# So the solution is either (6.66, 12) or (16,5). 

Answer 4

 

Let the two numbers be  x and y.

 

 3x + 4y = 68  (1)   "3 times one plus 4 times the other is 68"

  xy  = 80        (2)   "the product of the numbers is 80."

 

(2) implies  x = 80/y  and now "80/y" can be substituted for "x" in (1) to give:

 

  3(80/y) + 4y = 68

   240/y + 4y = 68     

   240 + 4y² = 68y     (multiplied both sides by y)

    4y² - 68y + 240 = 0    

    y² - 17y + 60 = 0   (divided both sides by 4)

    (y - 5)(y - 12) = 0

 

 y = 5   and  y = 12

 

y=5 —>   x=80/5 = 16

y=12 —>  x=80/12 = 20/3

 

 

Check:  (x=16, y=5)  3x + 4y = 3(16) + 4(5) = 48+20 = 68 (ok)

       (x=20/3, y=12)  3x + 4y = 3(20/3) + 4(12) = 20 + 48 = 68 (ok)

 

So there are two solutions:

     1.    x=16 and y=5

     2.    x=20/3 and y=12