Math help: Solving systems with 3 variables?

Question

I am soooo lost Please help!!!...This is a system of equations with 3 variables.
4x + 3z =4
2y - 6z =-1
3x +4y 3z =9
I have the answers just can't figure out how they got there.
(3/4,1/2,1/3)

Yest the last equation should be 8x+4y=3z =9
No wonder I couldn't come up with the correct answer. I will email the book publisher.

Answer 1

There's a typo in your last equation... if you plug in your values you don't get 9... The last equation should probably have an *8x* instead of 3x. If it is a typo in the book, that would explain it.

For their answers to work out, the 3 equations must be:
4x ....... + 3z = 4
........ 2y - 6z =-1
8x + 4y + 3z = 9

You can go about this in many ways, but basically you need to eliminate variables. For example, if you double the first equation, you can add it to the second to eliminate z.
8x ........ + 6z = 8
......... 2y - 6z = -1
8x + 2y = 7

Now subtract the 1st equation from the 3rd equation again eliminate z:
8x + 4y + 3z = 9
4x ....... + 3z = 4
4x + 4y = 5

Now you have two equations with just x and y:
8x + 2y = 7
4x + 4y = 5

If you double the first you'll be able to eliminate the y variable when you subtract:
16x + 4y = 14
4x + 4y = 5
12x = 9

Divide both sides by 12 and reduce:
x = 9/12
x = 3/4

Now just work backwards for the other variables. Start by plugging in x = 3/4 to one of the previous equations with x and y:
4x + 4y = 5
4(3/4) + 4y = 5
3 + 4y = 5
4y = 5 - 3
4y = 2
y = 2/4
y = 1/2

Finally substitute in x = 3/4 to the first equation to figure z:
4x + 3z = 4
4(3/4) + 3z = 4
3 + 3z = 4
3z = 4 - 3
3z = 1
z = 1/3

There's your answer which matches with the expected answer:
x = 3/4
y = 1/2
z = 1/3

Answer 2

4x + 3z = 4 ------------ eq1
2y – 6z = –1---------eq2
3x +4y 3z = 9---------eq3 ----------here sign of 3 z is not given taking this as + 3z
from eq1 and eq 2 eliminate out z term
multiply eq. 1 by 2 and add it with eq 1
8x + 6 z = 8----------- eq4 (eq 1 × 2)
8 x + 2y = 7 ---------eq 5(eq,4 + eq.1)
From eq. 1 value of 3z = 4 – 4 x substitute this value in eq 3
3 x + 4 y – 4x + 4 = 9
or – x + 4y = 5 ---------- eq 6
8 x + 2y = 7 ---------eq 5
–8 x + 32 y = 40 --------eq7-- (eq 6 × 8)
9 x = 1 ---------(eq 7 – eq 6)
x = 1/9

Answer 3

Are you particular you wrote this wisely? that is $6000 at a cost of 30$ AND $6000 at a cost of 40$???? besides, regardless of the rather gross sales is, that is what you do: Take the first pair of numbers, $6000 and $30, and plug them into the equation r = ap^2 + bp + c. you should get 6000 = 900a + 30b + c. Now do a similar utilizing the 2d pair of numbers. 6000 = 1600a + 40b + c. And the third: 5000 = 2500a + 50b +c. Now enable's sparkling up for a. Write the first equation in words of a (i.e. manage each and everything so as that in basic terms the 900a is on the right aspect). Then divide each and everything by technique of 900, and also you comprehend what a is. This value for a is expressed in words of b and c. Plug this value of a in to the 2d equation the position you be conscious the letter a. sparkling up this equation that now in basic terms incorporates the letters b and c for b. Now that you comprehend what b is the same as (in words of c in basic terms), you may plug this into the third equation, the position b is cutting-edge. be conscious that you nonetheless have a time period with a contained in the third equation, yet you already solved for a (in words of b and c) in equation a million. that is only a remember of substituting one variable in for yet another till you get all yet one to drop out of your equation.

Answer 4

Please check the equations. You have not typed correctly.

If the last equation is 8x + 4y + 3z = 9, then only the given

solution set is correct

Answer 5

use matrix to solve the problem by applying Gauss Elimination Method or Cramer's Rule.It is an easy one.you should be able to solve it within 5 to 10 minutes