Question

If 20 lb of rice and 10 lb of potatoes cost $16.20 and 30 lb of rice and 12 lb of potatoes cost $23.04, how much will 10 lb of rice and 50 lbs of potatoes cost?

Let x = price of rice per pound Let y = price of potaotes per pound 20x + 10y = 16.2 30x + 12y = 23.04 Solve the system of equations for x and y. Then substitute x and y in 10x + 50y and evaluate the expression.

20x+10y=16.2 R1
30x+50y=23.04 R2

Answer 1

Let x = price of rice per pound Let y = price of potaotes per pound 20x + 10y = 16.2 30x + 12y = 23.04 this is correct just solve for x and y

Answer 2

@precal thats where i am having problems.. i am not sure on how to solve for x and y.. the book has me confused with rows

Answer 3

the way to solve 2 equations with 2 unknowns is to "eliminate" one of the variables
20x+10y=16.2
30x+50y=23.04

Let's try to get rid of the y's
if the first equation had -50y (instead of 10y), when you added the 2 equations together, the -50y + 50y become 0 (and y is gone).

Notice that you can multiply both sides of the 1st equation by any number you want, and it will remain true. what number should you multiply it by to make the 10y become -50y ?

Answer 4

-5

Answer 5

yes, so multiply the 1st equation by -5
what do you get?

Answer 6

-5(20x+10y)=16.2 distribute -5 to both the 20 and the 10?
-50x-50y=16.2

Answer 7

yes, distribute the -5 and you have to multiply both sides

if we had (1+2)= 3 (which is true) and we multiply by -5
-5(1+2) = -5*3
-5*1 + -5*2 = -15
-5 -10 = =15
-15= -15 it is still true.

but if we do what you did
it would not work

Answer 8

thats not right..

Answer 9

what i had i mean

Answer 10

yes I know. start again: multiply both sides of the = by -5
when you distribute the -5 that means multiply *everything* inside the parens not just the y)

Answer 11

-100x-50y=-81

Answer 12

looks good. we now have
-100x-50y=-81
30x+50y=23.04

add the 2 equations

Answer 13

-70x+0y=-57.96 but that doesnt look right to me

Answer 14

0y is just 0
so you have
-70x = -57.96

divide both sides by -70

Answer 15

.828

Answer 16

Well, that was a nice exercise, but after you mentioned it didn't look right I re-read the question. the two equations we should start with are
20x + 10y = 16.2
30x + 12y = 23.04

Answer 17

Notice there isn't a nice number to multiply the first equation to get a -12y
(but there is a number, it is -12/10 or -1.2)

try that. multiply the first equation by -1.2

Answer 18

ok. -24x+-12y=-19.44

Answer 19

now add the two equations
-24x+-12y=-19.44
30x + 12y = 23.04

Answer 20

i got .6 or 3/5

Answer 21

x = price of rice per pound
so I would call it $0.60 (60 cents)

now use one of the 2 equations to find y
replace x with 0.6 and solve for y

It does not matter which, so use this one
20x+10y=16.2

Answer 22

.42

Answer 23

Finally,

how much will 10 lb of rice and 50 lbs of potatoes cost?

Answer 24

27 dollars

Answer 25

how did you get that?

Answer 26

i took my 10 pounds of rice and multiplied it by 60 cents and the 50 lbs of potatoes and multiplied by .42
then added the 2 together

Answer 27

yes, excellent.

Now here is a question. How did I pick -1.2 to multiply the first equation?
20x + 10y = 16.2
30x + 12y = 23.04

I saw the 10y and said, if that were 1y I could multiply by -12 and get -12y and -12y plus 12y in the other equation will add to 0.

then I said, if I divide by 10 then 10y will become 1y
so, putting those together I get -12/10 or -1.2 for the multiplier.

Can you pick a multiplier to get rid of the x's instead of the y's ?

Answer 28

3?

Answer 29

well a neg 3

Answer 30

-3 *20x = -60x and that does not exactly cancel out +30x in the other equation.
read my reasoning carefully.

Answer 31

-1.5?

Answer 32

yes. -1.5*20x= -30 x and added to +30x in the other equation gives 0!

I would get -1.5 by saying : divide 20x by 20 to get 1x. multiply 1x by -30 to get -30x

both steps together are : -30/20 or -1.5

Answer 33

Now you can solve 2 equations and 2 unknowns.

Answer 34

and that works for every problem?

Answer 35

i think you should rewrite my math book and have them publish it instead of what i have.. because you make so much more sense

Answer 36

yes, unless it's not solvable to begin with. (sometimes both equations are the same equation in disguise and that is a problem)
or sometimes the equations don't have a solution.

But if there is an answer, this way always works.

Answer 37

thanks.

Answer 38

is there an easy way to tell if it is not solvable

Answer 39

yes, sometimes when you eliminate x you will also eliminate y at the same time
and you get
0+0 = right side
if the right side is not 0, there is no solution. 0= number is never true.
if the right side is 0, then that means the 2 equations are really the same equation
and the solution is any x,y pair on the line (lots of solutions)

but in general, the two equations (there are lines) cross somewheres, and there is only one (x,y) pair that is on both lines (satisfy both equations)

Answer 40

and if 0=0
?

Answer 41

how can u do this problem on a graphing calc?

Answer 42

and if 0=0
?the solution is any x,y pair on the line (lots of solutions)

how can u do this problem on a graphing calc?

both equations are lines, and you graph them and see where they meet.

Here is your problem
http://www.wolframalpha.com/input/?i=20x+%2B+10y+%3D+16.2%2C++30x+%2B+12y+%3D+23.04

Answer 43

i have never heard of that website.. that helps alot

Answer 44

if you graph your equations, you will get 1 of 3 cases:
Case 1: lines will cross in one place, that is known as one solution
Case 2: lines are parallel and therefore do not cross (known as no solution)
Case 3: lines are the same, therefore cross in many places (known as many solutions)
hope that helps