SOLUTION: How do I solve the system of equations?
x+2y-6=z
3y-2z=7
4+3x=2y-5z
Algebra.Com
Question 176599: How do I solve the system of equations?
x+2y-6=z
3y-2z=7
4+3x=2y-5z
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
you have 3 equations in 3 unknowns.
you take 2 of the equations and eliminate one of the unknowns from it.
you take another 2 of the equations and eliminate the same unknown.
you are left with 2 equations in 2 unknowns.
you solve these 2 equations by eliminating one of the unknowns from them and solve for the remaining unknown.
once you have solved for one of the unknowns, you take one of the 2 equations in 2 unknowns and solve for the other unknown.
once you have solved for 2 unknowns, you take one of the original equations and solve for the 3d unknown.
once you have all 3 unknowns solved, you substitute in the original equations to see if you got it right.
---
you need to change the equations to standard form first.
standard form is:
ax + by + cz = k
where a,b,c are coefficients and k is a constant.
---
your first equation is:
x+2y-6=z
to convert this to standard form, do the following:
add 6 to both sides of the equation and subtract z from both sides of the equation.
the equation will become:
x + 2y - z = 6 (first equation)
---
your second equation is:
3y-2z=7
this is already in standard form.
3y - 2z = 7 (second equation)
---
your third equation is:
4+3x=2y-5z
to convert this to standard form, do the following:
subtract 4 from both sides of the equation and subtract 2y from both sides of the equation and add 5z to both sides of the equation.
the equation will become:
3x - 2y + 5z = -4 (third equation)
---
your 3 equations are now in standard form as follows:
x + 2y - z = 6 (first equation)
3y - 2z = 7 (second equation)
3x - 2y + 5z = -4 (third equation)
---
since one of the equations already has the x removed (it wasn't there to start with, you should take this as one of the equations you want to have with only 2 unknowns.
---
keep the second equation as is.
---
take the other 2 equations and multiply one or both by common factors to make the number of x's in both equations the same.
---
multiply the first equation by 3 to get the fourth equation as follows:
(x + 2y - z = 6) * 3 = 3x + 6y - 3z = 18 (fourth equation)
---
leave the third equation as is.
---
subtract the third equation from the fourth equation to get the fifth equation as follows:
(3x + 6y - 3z = 18) - (3x - 2y + 5z = -4) = (8y -8z = 22)
the fifth equation is:
8y - 8z = 22
---
your 2 equations with 2 unknowns are:
3y - 2z = 7 (second equation)
8y -8z = 22 (fifth equation)
---
you can eliminate the y or the z by multiplying by the right factors and then subtracting one equation from the other.
---
we will eliminate the z.
since -8z is 4 times -2z, we'll multiply the second equation by 4 to get the sixth equation and leave the fifth equation as is.
we get:
3y - 2z = 7 (second equation) * 4 = 12y - 8z = 28 (sixth equation)
8y -8z = 22 (fifth equation)
---
the two equations now become:
12y - 8z = 28 (sixth equation)
8y -8z = 22 (fifth equation)
---
we subtract the fifth equation from the sixth equation to get the seventh equation as follows:
(12y - 8z = 28) - (8y -8z = 22) = (4y = 6)
the seventh equation is:
4y = 6
---
divide both sides of this equation to get:
y = 6/3 = 3/2
---
you now have one of the unknowns:
y = 3/2
substitute this in the second or fifth equation and solve for z.
we'll use the fifth equation as follows:
---
fifth equation is:
8y -8z = 22
substitute 3/2 for y to get:
8*3/2 - 8z = 22
multiply both sides of equation by 2 to get:
8*3 - 2*8*z = 2*22 which becomes:
24 - 16z = 44
subtract 24 from both sides of this equation to get:
-16z = 44-24 = 20
divide both sides of this equation by -16 to get:
z = -20/16 = -5/4
---
you now have 2 of the unknowns.
you have:
y = 3/2
z = -5/4
---
substitute in first or third equation to solve for x.
we'll use first equation as follows:
first equation is:
x + 2y - z = 6
substitute 3/2 for y and -5/4 for z to get:
x + 2*(3/2) - (-5/4) = 6
multipy both sides of this equation by 4 to get:
4x + 4*2*(3/2) - 4*(-5/4) = 4*6
which becomes:
4x + 12 - (-5)) = 24
which becomes:
4x + 12 + 5 = 24
which becomes:
4x + 17 = 24
subtract 17 from both sides to get:
4x = 24-17 = 7
divide both sides by 4 to get:
x = 7/4
---
you now have all 3 of the unknowns.
they are:
x = 7/4
y = 3/2
z = -5/4
---
substitute in first equation as follows:
first equation is:
x + 2y - z = 6
this becomes:
7/4 + 2*(3/2) - (-5/4) = 6
simplify to get:
7/4 + 6/2 + 5/4 = 6
multiply both sides by 4 to get:
7 + 12 + 5 = 24
combine like terms to get:
24 = 24
first equation is true so value are good for first equation.
---
substitute values in second equation as follows:
second equation is:
3y - 2z = 7
this becomes:
3*(3/2) - 2*(-5/4) = 7
which becomes:
9/2 - (-10/4) = 7
which becomes:
9/2 + 10/4 = 7
multiply both sides of this equation by 4 to get:
18 + 10 = 28
which becomes:
28 = 28
second equation is true so values are good for second equation.
---
substitute values in third equation as follows:
third equation is:
3x - 2y + 5z = -4
this becomes:
3*(7/4) - 2*(3/2) + 5*(-5/4) = -4
which becomes:
21/4 - 6/2 + (-25/4) = -4
multiply both sides of this equation by 4 to get:
21 - 12 - 25 = -16
combine like terms to get:
-16 = -16
third equation is true so values are good for third equation.
---
values are good for all equations.
answer is:
x = 7/4
y = 3/2
z = -5/4
---
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