Problem 10 can be solved with a system of linear equations.
Let x= the number of students a van can hold, and let y= the number students a bus can hold.
Your first equation is 372=x+6y, because 372 students equals however many students were in a van, plus how many students were in 6 buses.
Your second equation is 780=4x+12y, because 780 students equals however many students were in 4 vans, plus how many students were in 6 buses.
You can solve this system a number of ways, I chose elimination.
First, you multiply your first equation by -2, that way your y's can cancel each other.
Now add the equations together, and you should get 36=2x, divide both sides by 2, and you get 18=x, which means that each van can hold 18 students.
Now plug your x variable into either equation, I plugged it into the first equation.
372=(18)+6y. Subtract 18 from each side, and you get 354=6y. Divide each side by 6 to get 59=y, which means that each bus can hold 59 students.
You can check these answers by plugging them back into the equations.
372=(18)+6(59)
372=18+354
372=372
Plugging the answers into the other equation results in a true statement as well. My only question is what kind of buses and vans the schools are taking to fit that many kids... lol.
Question 11 is similar to 10.
Let the cost of the small orange boxes equal x, and the cost of the large orange boxes equal y.
Your first equation will be 203=3x+14y, because 203 dollars equals 3 times the cost of a small box, plus 14 times the cost of a large orange box.
Your second equation will be 220=11x+11y.
Once again, you can solve this a number of ways, I chose substitution.
First get a variable by itself on one equation, I chose x on the 1st equation.
203=3x+14y
-14y -14y
203-14y=3x
[tex] \frac{203-14y}{3} [/tex]
[tex] \frac{203}{3}- \frac{14y}{3} =x[/tex]
Now plug x into the other equation.
[tex]220=11y+11(\frac{203}{3}- \frac{14y}{3})[/tex]
[tex]220=11y+ \frac{2233}{3} - \frac{154y}{3} [/tex]
[tex]220= \frac{-121y}{3} + \frac{2233}{3} [/tex]
[tex] \frac{-1573}{3}= \frac{-121y}{3} [/tex]
[tex]-1573=-121y[/tex]
[tex]13=y[/tex]
Whew, that was messy. But we now we know that each large orange box costs $13. Now, we can plug y into an equation to solve for x. I'm going to plug it into the 2nd equation.
220=11x+11(13)
220=11x+143
77=11x
7=x
So each small orange box costs $7.
Plugging your answers into either equation will prove the equation to be true.
Question 12 is another system of equations.
Let the number of press model A = x, and the number of press model B = y.
x+y=14 is the first equation, because however many of press A, plus however many of press B, equals 14 presses in total.
70x+55y=905 is the 2nd equation, because 70 times however many model A presses there are, plus 55 times however many model B presses there are, equal the number of books they produce a day.
I solved this equation with substitution.
x+y=14
x=14-y
Now we plug x into the other equation.
70(14-y)+55y=905
980-70y+55y=905
980-15y=905
-15y=-75
y=5
So they have 5 model B presses. Now you can plug in y to solve x. I plugged y into the 1st equation.
x+5=14
x=9
So they have 9 model A presses. Once again, you can plug in your answers into either equation and it will be true.
Sorry for such a long answer, if you have any more questions, feel free to ask me.
