Question

an inventor bought 600 shares of stock, some at $6.25 per share and some at 1.00 per share. If the total cost was $1387.50, how many shares of each stock did the investor buy? (round to two decimal places if necessary)

Answer 1

call the shares bought at $6.25 \(x\) and the shares bought at $1 \(y\)

\[x+y = 600\]\[6.25x+1y=1387.50\]Solve the system of equations to find \(x\) and \(y\). Do you know how?

Answer 2

I am not sure because I dont know where to start. Could you walk me through it please?

Answer 3

do u mean u need help to solve the system of equation ?

Answer 4

yes

Answer 5

Pick one of the variables in the first equation and solve for it. You'll get either \[x = 600-y\]or\[y=600-x\]Now replace the variable on the left side with the expression on the right side wherever you see that variable in the second equation. That gives you an equation in terms of one variable, which you should be able to solve without assistance. Find the value of that variable, then plug it into the first equation to find the value of the other variable.

Answer 6

That's called the substitution method. Another method is called elimination, but it's time for me to call it a night, maybe Isuru will explain it to you.

Answer 7

okay then thank you

Answer 8

were you able to solve the problem?

Answer 9

no I wasnt. Im confused of where to start

Answer 10

Pick a letter, \(x\) or \(y\)

Answer 11

Well?

Answer 12

I picked y and I got the answer x=162 and y=-102?

Answer 13

\[x+y = 600\]Solve for \(y\):\[x-x+y=600-x\]\[y=600-x\]Did you get that?

Answer 14

yeah I did.

Answer 15

Okay. What did you get when you substituted that into the other equation?

Answer 16

I got 3750-6.25x=1387.50. I got -378?

Answer 17

How the heck did you get that?

\[6.25x + 1y = 1387.50\]Right?

Answer 18

Now replace \(y\) with \((600-x)\), what do you get?

Answer 19

\[6.25x + 1(600-x) = 1387.50\]That's the result of the substitution of \(y = 600-x\)
Solve that for \(x\), please.

Answer 20

I got 150?

Answer 21

6.25 per share =x=150

Answer 22

How many shares at the $1 price?

Answer 23

I got y=450?

Answer 24

Good. Let's check the result:

150 shares @ 6.25 + 450 @ 1
150*6.25 + 450*1 = 937.5 + 450 = 1387.50
that part checks
150 + 450 = 600
that part also checks

It's important to test the solution in all of the original equations, because it is possible to construct erroneous solutions which will satisfy some of the original equations, but not all. For example, 500 of the cheap stock and 100 of the expensive gives you 600 shares, but won't satisfy the investment pricing equation. We could also make up a set of numbers that satisfies the pricing equation, but doesn't satisfy the share count.

Answer 25

I suppose I could show you the elimination technique if you are interested.

Answer 26

Okay then. The more I learn, the better

Answer 27

Good answer! :-)

So with elimination, we want to find a set of operations we can do to combine the equations in such a fashion that we eliminate variables until we are left with an equation in only one variable to solve. Then we use that variable to find the values of the other variable(s), much like we did in substitution.

\[x+y=600\]\[6.25x+1y=1387.50\]

We would like to multiply one or both of the equations by numbers chosen so that we end up with one of the variables having coefficients which are equal in magnitude but opposite in sign. For example, in this problem, we've got \(y\) in the first equation and \(1y\) in the second. If we multiply one of the equations by -1, we'll have a \(1y\) and a \(-1y\), and if we add the equations together, that will give us \(0y\), leaving us an equation just in terms of \(x\). Does that make sense so far?

Answer 28

yes it does.

Answer 29

Multiplying the whole equation by the same number doesn't change the solution at all because we are doing the same thing to both sides of the equation. 2x=4 has the same solution as 2*2x = 4*2, right?

Answer 30

okay

Answer 31

Okay, so let's do just that:

\[-1*x -1*y = -1*600\]\[6.25x+1y=1387.50\]
-------------------------------
\[6.25x-1x + 1y-1y = 1387.50-600\]\[5.25x=787.50\]Solve that for \(x\), then plug the value into either of the original equations to find \(y\).

Answer 32

Here it was easy because we already had a variable where the coefficients were equal. We could have just subtracted one equation from the other instead of multiplying by -1 and adding. Sometimes you'll have a more complicated setup, like this:
\[2x+3y=6\]\[3x+4y=0\]
Here I would multiply the first equation by -3, and the second by 2:
\[-3*2x-3*3y=-3*6\]\[2*3x+2*4y=2*0\]
\[-6x-9y=-18\]\[6x+8y=0\]--------------\[0x-y=-18\]\[y=18\]\[2x+3(18)=6\]\[2x+54=6\]\[x=-24\]

Answer 33

It's a lot like making a common denominator when adding fractions:

\[\frac{1}{2} +\frac{1}{3} = \frac{1}{2}*\frac{3}{3} + \frac{1}{3}*\frac{2}{2}\]

Answer 34

okay that makes sense.

Answer 35

just multiply by the coefficient from the other equation, changing the sign if necessary.

sometimes you get lucky and the coefficient in one equation is a multiple of that in the other equation, in which case you only have to multiply one equation by something.

Answer 36

so that's elimination. you can do it on larger problems, like 3 equations in 3 unknowns. you just have to repeat the process to get rid of each additional variable. if you have 3 eqs in 3 unks, you would first eliminate a variable from the first pair of equations (1st and 2nd), and then the same variable from the second pair of equations (2nd and 3rd). Now you have two new equations in only two unknowns, and you solve them like we did here. With those solved, you go back and find the third variable's value.

Answer 37

Thank You so much for helping me understand all of this. you are awesome!!

Answer 38

You can do larger systems with substitution as well, but the algebra gets messier.

Answer 39

Yet another way to solve a system of equations like this is by graphing. If the two lines intersect, the coordinates of the point at which they intersect is the solution. Here's a graph showing that for this problem.

Answer 40

If they don't intersect (because they are parallel), there is no solution. If the lines are identical, there are an infinite number of solutions.

When you solve a system where there is no solution, you'll end up with something crazy like \[4=0\]A false result like that means no solution, because the lines are parallel.

If you instead get something like \[0=0\]that means the lines are coincident (the same) and there are infinitely many solutions. Any point on the line represents a solution in this case.