Question

A truckload of 10-pound and 50-pound bags of fertilizer weighs 9000 pounds. A second truck carries twice as many 10-pound bags and half as many 50-pound bags as the first truck. That load also weighs 9000 pounds. How many of each bag are on the first truck?

Answer 1

Answer:

300 10-pound bags and 120 50-pound  bags.

Step-by-step explanation:

To solve this problem, we need to use Solving Linear Systems by multiplying first method.

We need to create an equation for each truck that we have.

1st truck can be shown as:

$x+y=9000$

2nd truck can be shown as:

$2x+\dfrac{1}{2}y=9000$

We use 2x because the amount the second truck carries is TWICE the amount of the first truck, and 1/2y because it has HALF as many as the first truck.

Now that we have both our equations, let's begin by putting them together

$x+y=9000$

$2x+\dfrac{1}{2}y=9000$

Now we can see that we have a fraction for our second equation, let's remove that fraction by multiplying the entire equation by 2.

$x+y=9000$

$2x+\dfrac{1}{2}y=9000$

$2(2x+\frac{1}{2}y=9000)$

The new equations will be:

$x+y=9000$

$4x+y=18000$

Now that we have both equations with whole numbers we can then proceed to +/- the equations to find a value of x or y by eliminating either of the two.

Let's remove the y by subtracting both equations to solve for x.

x  + y = 9000

-4x+y = 18000

$\dfrac{-3x}{-3}=\dfrac{-9000}{-3}$

x = 3000

Now that we have a value for x, we can then substitute x from any of the two equations.

Let's use the equation $x+y=9000$:

$x+y=9000$

$3000+y=9000$

$y=9000-3000$

y = 6000

Now that we found both values of x and y, we can then simply divide the total values of each from the weight of the bags to find how many of each bag are present in the first truck.

x = 3000 (10-pound bags)

3000 / 10 = 300 10-pound bags

y = 6000 (50-pound bags)

6000/50 = 120 50-pound bags