Okay, here's a question. What are the steps to figure out two variables with two equations? For example: 3x+2y=36 4x-y=6 What are x and y so that both equations are true? Anybody?

to do this problem somehow the middle 2y needs to negate itself inorder to get rid of the 2y we need to multiply the second part 4x-y=6 by 2: 2(4x) - 2(y) = 2(6) which gives you 8x-2y=12, now line the problem up and subtract 3x+2y=36 -8x-2y=-12 ---------- -5x = 24 x= 24/-5 = -4.8 plug that back in to get y (you can plug it into either equation) 3(-4.8)+2y=36 -14.4 +2y=36 2y=50.4 y=25.2 then y=25.2, x= -4.8 is plugged back into the original: 3x+2y=36 4x-y=6 and presto!

That's right! I knew there was a simple step I was forgetting (canceling one variable). Thanks! But are you sure you worked it out right? To get rid of the second equation, you multiply it by 2. I get that part. So you'd get: 8x - 2y = 12, but if you SUBTRACT that from the first equation, wouldn't you end up with -5x + 4y = 24 since the y in the second equation is negative?

i think i figured it out - it says you can ADD or SUBTRACT to eliminate so after i multiplied by 2, I see that the middle numbers eliminate themselves if i add... 3x+2y=36 8x-2y=12 ---------- 11x=48 x=4.36363636 substituting 4.36 repeating y = 11.454545 the only problem is when i put x and y back into original and i got a perfect 6 but 35.99999 ( i think that counts as 36, right?)

That is right, Eki. I did it an easier way. At least I think it is. Multiply both equations so you can cancel out the x's. Multiply top one by 4 and bottom one by 3. 4(3x + 2y) = (36)4 so 12x + 8y = 144 3(4x - y) = (16)3 so 12x - 3y = 18 Now subtract You get 11y = 126 y = 126/11 or 11.4545 Substitute for y and solve for x 4x - 11.4545 = 6 4x = 17.4545 x = 17.4545/4 So, x = 17.4545/4 and y = 126/11 I used a common denominator of 44 then substituted into the second equation and it worked. 4(192/44) - 504/44 = 6 192/11 - 126/11 = 66/11 = 6

these are the instructions for solving two unknowns - systems of equations from cliffs.... 1. multiply one or both equations by some number to make the number in front of one of the letters (unknown) the same in each equation 2. add or subtract the two equations to eliminate one letter 3. solve for the other unknown 4. insert value of the first unknown in one of the original equations to solve for the second unknown.. a unsolvable case is: 3a+4b=2 6a+8b=4 the second equation is actually the first equation multiplied by 2 and in this instance, the system is unsolvable

hi! i think we are both working off the same principle - you multiplied both equations by a number in order to eliminate a variable. I only multiplied one equation by one number in order to eliminate it.. i hope that the actual exam will use easy equations with even divisible numbers -- fingers crossed!!!

Thanks for the reminder about how to do these problems. It's been a while. I doubt that there will be problems like mine on the CSET with numbers that aren't evenly divisible. I just picked random numbers for my question. lol. Thanks again!

LOL back, Eki - you are a stinker! Upsadaisy and Jello illustrate one of the Great Unspoken Truths of Math: There may indeed be only one answer, but there's rarely only one right way to get to it. And in fact this truth is why elementary teachers DO need to know math and to know it to a certain level of comfort. You need to know math for the times when the teacher's manual gives Upsadaisy's solution but your kids hand in Jello's solution or still other versions we haven't seen here, all of which really do work. Sure, it's easier to mark the answer wrong because it isn't what the book says - but it's much cooler to be able to work your way through the answer, find that it works, and know that a kid in your class knows the principles that well AND you had a hand in that. "Yeah," somebody says, "but this is algebra. That isn't going to happen in the primary grades." Oh, yes, it will - at least, I hope it will. Little kids can do astonishing things in math, at least until somebody shames it out of them. Please find books and look for Web sites that help you see that math really is very, very cool. One fine book - perhaps my favorite from Barron's - is MATH WIZARDRY FOR KIDS.

So, true, TeacherGroupie. And it's what math so much fun (I know some wouldn't agree with me on that one!) I am the math coordinator for our school and have been asked to do lots more next year. My first big job was to make up a list of all the items we need to order - my favorite type of job. I hope they follow through because we have nothing and most of the teachers don't augment with own funds - because our salaries are so pitiful. In fact, it is going to be harder to motivate them than the students. Anyway, I teach 5th self-contained and also one middle school class per day. Next year will be 8th. I'll have most of the same kids I had this year. I will be tutoring this summer - and will even have a former student now going into 11th grade who wants to brush up on algebra I before going into algebra II. Have you used Hands On Equations?

It's a hands-on approach to algebra. Kids as young as 3rd grade can quickly be solving equations such as this: 2x + 3(4x + 5) = 2x + 1 . My 5th graders learned to solve ones such as that in 3 days. It is wonderful and I think, when started young, gives them an internalized understanding of balance, maintaining equality when using same operations on both sides of equations. It can be used for all ages, even adults. It is very simple to teach and to use. I have only looked at Alge Lab and it seems a bit confusing. I love Hands On Equations. Take a look at their website.