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# I need problems 1-3 on page 157. The ones on solving by elimination for (x,y,z)?

Topic: Math online answers for homework
July 16, 2019 / By Burgundy
Question: If you have this can you get them for me w/ answers cause i suck in math nd its my homework. i forgot the stupid book at school and i can't acess it online cause i don't have a webcode. Prentice Hall or Pearson Algebra 2 copyright 2007 I'min 11th grade. This is the book. Algebra 2 (2007 Edition) Prentice Hall PLEASE HELP OR I'LL FAIL THE QUARTER

## Best Answers: I need problems 1-3 on page 157. The ones on solving by elimination for (x,y,z)?

Allissa | 1 day ago
specific i will believe it. it is how one might circulate approximately simplifying this equation. 5(-3x-2)-(x-3)= -4(4x+5)+a million (-15x-10)-x+3= (-16x-20)+a million -15x-10-x+3= -16x-20+a million -15x-x+16x= -20-3+a million+10 -16x+16x= -23+11 0 =/= -12 Yep, constructive that it is -12 (neg. twelve) . a million) Multiply 5 by using each and everything interior the parentheses. circulate around the (=) sign and multiply -4 by using each and everything in that parentheses. 2) clean the parentheses 3) it is the equation devoid of () 4) team like numbers and signs and warning signs (x's with x's) (numbers w/ #) 5) Simplify answer 0 isn't equivalent to a -12
👍 250 | 👎 1
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We found more questions related to the topic: Math online answers for homework

Originally Answered: Solving Systems With Elimination?
By elimination: Multiply Equation 1 and 2 by -1: -3a + 3b - 4c = 23 -a - 2b + 3c = -25 4a - b + c = 25 Add the equations together. You are left with: 0 = 23 Obviously, that is impossible, so there is NO solution for this system of equations. ---------------------------------------... By matrix: 3 -3 4 | -23 1 2 -3 | 25 4 -1 1 | 25 Switch Row1 with Row2: 1 2 -3 | 25 3 -3 4 | -23 4 -1 1 | 25 Let Row2 = Row2 - 3*Row1: 1 2 -3 | 25 0 -9 13 | -98 4 -1 1 | 25 Let Row3 = Row3 - 4*Row1: 1 2 -3 | 25 0 -9 13 | -98 0 -9 13 | -75 At this point, you can see there is going to be a problem. Rows 2 and 3 are identical on the left side of the equation, yet they have different answers. That means there will be NO solution for this set of equations. =======================================...

Uziah
How can we solve the problem when you dont have one? You need to remember to bring your books home to do your homework. I have not been able to read a math problem without the text book or problem written down on paper.
👍 110 | 👎 -3

Rohan
It might help if you specify which book it is, people can't really help unless they know the exact book you need the problems from.
👍 110 | 👎 -7

Milburn
Whoa, whoa, whoa. What book is this? What math level? What school? No one's gonna give you an answer until you answer this other stuff, and even then, you probably won't get an answer.
👍 110 | 👎 -11

Originally Answered: Sense nobody answered the 1st time.solving by elimination algebra problem?
Answers only work once so learning methods is recommended. 10x + 2y + z = 885 5x + 4y + z = 865 9x + 6y + 6z = 1410 First aim was to eliminate (get rid of) 2 variables to find x. You were on the right track but made a small error. 2) Subtract the 2nd from equation from 1st. 5x - 2y =20 (notice the minus)............(a) 3) Need another equation with just x and y, so multiply 2nd by 6 so it has same number of z's as 3rd 30x + 24y + 6z = 6*865 9x + 6y + 6z = 1410 now subtract (same as *-1 and add) 21x + 18y = 3780..............................(b) 4) (a) and (b) form a linear system in two variables. Solve this system for x. To do that multiply (a) by 9 so number of y's match up 45x - 18y = 180...........................(c) and add to (b) 21x + 18y = 3780..............................(b) 66x = 3960 so x = 60 5) Substitute the value of x into the first 2 equations. 2y + z = 885 - 600 = 285 4y + z = 865 - 300 = 565 6) Solve the system in 5) for y and z by subtracting 2y = 565 - 285 = 280, so y = 140 z = 285 - 280 = 5 7) Write the cost of each item. Sleeping bags \$60, tents\$140, bug repellent \$5 This sort of question is likely to be in an exam so read it all through again until you are sure you no longer confused and could do this without help. Regards - Ian

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