Can you please help me do the following problem
2x +2z=6
5x+3y =11
3y - 4z=1
can you please explain to me how to do it the answers in the back in the book are (-2,7,5). I keep trying the problem but I keep getting the wrong answer can you please help.
2007-01-11 15:05:20 · 5 answers · asked by elizabeth g 2 in Science & Mathematics ➔ Mathematics
Thanks you guys-you guys are awesome..really helped
2007-01-11 15:27:36 · update #1
First solve the first equation for z:
2z= -2x + 6
z = -x + 3
Second substitute this into the third equation:
3y - 4(-x + 3) = 1
3y + 4x - 12 = 1
3y + 4x = 13
Now solve this new equation with the second equation using Elimination method. Subtract the equations
5x + 3y = 11
4x + 3y = 13
Leads to x = -2
Sub this into the other equations:
z = -(-2) + 3 = 2 + 3 = 5
5(-2) + 3y =11
3y = 11 + 10
3y = 21
y = 7
Therefore the solution is (-2, 7, 5).
2007-01-11 15:13:02 · answer #1 · answered by keely_66 3 · 1⤊ 0⤋
OK. I'll help, but this has nothing to do with Trig.
2x +2z=6 <--- Eq1
5x+3y =11 <--- Eq 2
3y - 4z=1 <--- Eq 3
Multiply Eq 1 by 2 getting:
4x+4z =6 <--- Eq4
Add Eq 3 to Eq 4 getting:
4x +3y = 7 <--- Eq 5
Subtract Eq 5 from Eq 2 getting:
x = 4
Put x =4 in Eq 5 getting:
4*4 +3y =7
3y = 7 -16
3y =-9
y = -3
Put x=4 into Eq 1 getting:
2*4 +2z=6
2z = 6-8 = -2
z = -1
Summary x=4, y=-3, z = -1
2007-01-11 23:21:31 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
First, number your equations from top to bottom 1, 2, and 3, respectively.
1. Solve equation 1 for x in terms of z:
2x+2z=6
2x=6-2z
x=3-z
2. Then solve equation 2 for y in terms of z (by substituting the equation above for x):
5x+3y=11
5(3-z)+3y=11
15-5z+3y=11
-5z+3y=-4
3y=-4+5z
y=(5z-4)/3
3. Then solve problem 3 in terms of z by substituting y for the equation above:
3y-4z=1
3[(5z-4)/3]-4z=1
5z-4-4z=1
z-4=1
z=5
4. Now that you know the solution to z, go back and solve equation 1 for x:
2x+2z=6
2x+2(5)=6
2x+10=6
2x=-4
x=-2
5. Now that you know the solution to x, go back and solve equation 2 for y:
5x+3y=11
5(-2)+3y=11
-10+3y=11
3y=21
y=7
Therefore, x=-2, y=7, and z=5.
2007-01-12 00:12:09 · answer #3 · answered by nubianwargod 1 · 0⤊ 0⤋
i haven't done trig since the early 90s but i think you solve for x first.
using 2x + 2x = 6
subtract 2z from both sides of the equation and you get 2x = 6 - 2z
divide each side by 2 and you get
x = 3-z
so now you can use 3-z instead of x in the other equations
and so on....
2007-01-11 23:15:35 · answer #4 · answered by Laura W 2 · 0⤊ 0⤋
2x +2z=6 ........(1)
5x+3y =11 ......(2)
3y - 4z=1 ........(3)
(2)and(3)
5x + 3y = 11
3y - 4z = 1
----------------- -
5x + 4z = 10 ......(4)
(1)and(4)
2x + 2z = 6 | times by 5
5x + 4z = 10 | times by 2
--------------------------------
10x + 10z = 30
10x + 8z = 20
-------------------- -
2z = 10
z = 5
2x + 2z = 6
2x + 2 . 5 = 6
2x + 10 = 6
2x = -4
x = -2
5x + 3y = 11
5 . (-2) + 3y = 11
-10 + 3y = 11
3y = 21
y = 7
(x,y,z) = (-2,7,5)
hope help you.
Solved!!!
2007-01-11 23:15:57 · answer #5 · answered by fortman 3 · 0⤊ 0⤋