Problem: 3rd edition, slightly different from 2nd edition: In this edition, the total weight of the 3 beans must add to 14,000. The owner of the cafe can only spend $14,800. The beans cost $0.8, $1.2, and $1.8. Also, different from the 2nd editon, The weight of the least expensive bean must be equal to the total weight of the middle grade bean plus twice the weight of the most expensive bean.
Let x = the weight of the 0.8 bean, the least expensive
y = the weight of the middle $1.2 bean
z = the weight of the most expensive bean
Here are the 3 equations you can build:
x + y + z = 14,000 (the total weight of the beans is 14,000)
.8x + 1.2y + 1.8z = 14,800 (the total cost of the beans is $14,800)
x = y + 2z (the cheapest bean weight is equal to the middle bean plus twice the most expensive)
Notice that the last equation is already solved for x. We will start there. If we did not have one of the variables solved in any of the 3 equations, we would need to do that first. Now that we know what x is, we substitute it in for x in the other 2 equations:
(y + 2z) + y + z = 14,000
.8(y + 2z) + 1.2y + 1.8z = 14,800
Simplify the first equation by combining y and z
2y + 3z = 14,000
Simplify the 2nd equation:
.8y + 1.6z + 1.2y + 1.8z = 14,800
2y + 3.4z = 14,800
We now have 2 equations:
2y + 3z = 14,000
2y + 3.4z = 14,800
The problems is now just like the ones we did in Chapter 1.10. Solve for one of the variables. We have 4 choices. Solve for y or z in equation 1 or solve for y or z in equation 2. Pick the easiest for you. I will pick solving for y i the first equation:
2y + 3z = 14,000
2y = 14,000 – 3z
y = (14,000 – 3z) / 2
y = 7000 – (3/2)z (distributive prop.. mult. by 1/2)
One more step.. substitute this value for y into the y in the last equation, in our case, equation 2: 2y + 3.4z = 14,800
2(7000 – (3/2)z) + 3.4z = 14,800
14,000 – 3z + 3.4 z = 14,800
14,000 + .4z = 14,800
.4z = 800
z = 2000 (divide both sides by .4)
Now we know z. You cannot find x right now, since the equations with x in them also have both a y and a z. Not yet solvable. But we do have 2 equations with only a y and a z. We now know z, so we can solve for y.
2y + 3z = 14,000
2y + 3.4z = 14,800
I will pick the first one. Using the 2nd will get the same answer
2y + 3(2000) = 14,000
2y + 6000 = 14,000
2y = 8000
y = 4000
So you now know that you have 2000 pounds of the most expensive bean z, and 4000 pounds of the middle grade bean. Use these 2 values in any of the 3 original equations to solve for x, the weight of the cheapest bean..
x + y + z = 14,000
x + 4000 + 2000 = 14,000
x + 6000 = 14,000
x = 8000
So, you would buy 8000 of the x or cheapest bean, 4000 of the middle grade bean, and 2000 of the most expensive bean to satisfy your conditions.