36x-15y+50z=-10
2x+25y=40
54x-5y+30z=-160
are you solving for x, y, and z ?
what method have you learned ?
I would change the second equation to
2x = 40 - 25y
now think of the first equation,
36x-15y+50z=-10 as
18(2x) - 15y + 50z = -10
18(40 - 25y) - 15y + 50z = -10
simplify and do the same thing to the third equation.
Now you have 2 equations in y and z, which can be readily solved.
Let me know what you got.
The method we learned is like multiplying an equation and then add it to another equation to cancel a variable. I did:
36x-15y+50z=-10
-162x+15y-90z=480 (3rd eq. multiplied by 3)
which gave me -126x-40z=470
then I did:
2x+25y=40
270x-25y+150z=-800 (3rd eq. multiplied by 5)
which gave me 272x+150z=-760
This is the part where I'm stuck on because I usually would combine the 2 new equations (after multiplying one of them to cancel out a variable) to cancel out a variable. But in this case I can't find a number that helps?
ok, I would now work on the z's
you have -40z and +150z
the LCM of 40 and 150 is 600, so multiply your -126x-40z=470 by 15 and
272x+150z=-760 by 4
You would then add to eliminate the z's
I did not check your arithmetic, so good luck
Oh okay I forgot to use the LCM. Thanks :]
To solve this system of equations, we can use the method of elimination or substitution. Let's use the method of elimination.
1. Write down the three equations:
36x - 15y + 50z = -10 (Equation 1)
2x + 25y = 40 (Equation 2)
54x - 5y + 30z = -160 (Equation 3)
2. Choose two equations to begin eliminating a variable. In this case, let's eliminate the y-variable.
Multiply Equation 2 by 15 to make the coefficients of y in both equations equal:
30x + 375y = 600 (Equation 4)
3. Multiply Equation 1 by 25 to make the coefficients of y in both equations equal:
900x - 375y + 1250z = -250 (Equation 5)
4. Subtract Equation 4 from Equation 5 to eliminate the y-variable:
(900x - 375y + 1250z) - (30x + 375y) = -250 - 600
870x + 1250z = -850 (Equation 6)
5. Now, let's eliminate another variable. In this case, let's eliminate the z-variable.
Multiply Equation 2 by -15 to make the coefficients of z in both equations equal:
-30x - 375y = -600 (Equation 7)
6. Multiply Equation 3 by 10 to make the coefficients of z in both equations equal:
540x - 50y + 300z = -1600 (Equation 8)
7. Add Equation 7 to Equation 8 to eliminate the z-variable:
(540x - 50y + 300z) + (-30x - 375y) = -1600 - 600
510x - 425y = -2200 (Equation 9)
8. Now we have two equations with two variables:
870x + 1250z = -850 (Equation 6)
510x - 425y = -2200 (Equation 9)
9. We can solve this system of equations by either substitution or elimination.
Option 1: Elimination
Multiply Equation 6 by 425 and Equation 9 by 250 to cancel out the coefficients of x:
369750x + 531250z = -361250 (Equation 10)
127500x - 106250y = -550000 (Equation 11)
Multiply Equation 10 by 2 and Equation 11 by 3 to make the coefficients of z in both equations equal:
739500x + 1062500z = -722500 (Equation 12)
382500x - 318750y = -1650000 (Equation 13)
Subtract Equation 12 from Equation 13 to eliminate the z-variable:
(382500x - 318750y) - (739500x + 1062500z) = (-1650000) - (-722500)
-357000x - 1593750z = -927500
Now we have two new equations:
-357000x - 1593750z = -927500 (Equation 14)
413750x - 106250y = -927500 (Equation 15)
Multiply Equation 14 by -413750 and Equation 15 by 357000 to cancel out the coefficients of x:
147752500000x + 659421875000z = 382418750000 (Equation 16)
147757500000x - 56427500000y = -330322500000 (Equation 17)
Subtract Equation 16 from Equation 17 to eliminate the x-variable:
(147757500000x - 56427500000y) - (147752500000x + 659421875000z) = (-330322500000) - 382418750000
499500000z = -712741250000
Solve for z:
z = -712741250000 / 499500000 = -1425
Substitute z = -1425 back into Equation 6 or 9 to find x or y.
Option 2: Substitution
Solve Equation 6 or 9 for x or y and substitute it into the other equation to find the remaining variable.
Once you have determined values for x, y, and z, you can substitute them back into any of the original equations to check if they satisfy the system.
Please note that this solution process involves a lot of steps, so it may be easier to use matrix methods or software to solve such systems of equations.