Solve the System Equation?

Question

x + y + z = 2
3x - y + z = 4
2x + y + 2z = 3

Answer 1

z= 2 - x - y

3x - y + 2 - x - y=4
2x - 2y - 2 = 0 => x - y -1 = 0

2x + y + 4 - 2x -2y = 3
-y = -1 => y = 1

x = 1 + y = 2

z = 2 - x - y = 2 - 2 - 1 = -1

Answer 2

x + y + z = 2
3x - y + z = 4
2x + y + 2z = 3

To solve this, express one variable in terms of the other two in the first equation, and plug into the 2nd and 3rd equations.

x + y + z = 2 implies x = 2 - y - z. Plugging this into the other two,

3(2 - y - z) - y + z = 4
2(2 - y - z) + y + 2z = 3

Simplify this to obtain two equations and two unknowns. Remember to move all constants (numbers without letters) to the right hand side.

6 - 3y - 3z - y + z = 4
4 - 2y - 2z + y + 2z = 3

-3y - 3z - y + z = -2
-2y - 2z + y + 2z = -1

Our two equations/two unknowns are then:

-4y - 2z = -2
-y = -1

We can immediately see that y = 1. Plugging y = 1 into the first equation, we get -4(1) - 2z = -2, -4 - 2z = -2, -2z = 2, z = -1.

y = 1, z = -1. We can now get the third variable, x, easily.

x = 2 - y - z, so x = 2 - 1 - (-1) = 2 - 1 + 1 = 2

x = 2, y = 1, z = -1.

It's a good idea to check our solutions to see if they work (since any arithmetic error may influence the final result).

Plugging x = 2, y = 1, z = -1 into the first equation,

x + y + z = 2 + 1 + (-1) = 2, so this works.
3x - y + z = 3(2) - (1) + (-1) = 6 - 1 - 1 = 4, so this works.
2x + y + 2z = 2(2) + 1 + 2(-1) = 4 + 1 - 2 = 3, so this works.

The obtained results work!

x = 2
y = 1
z = -1

Answer 3

i duno hw to write out the step but according to calculator casio 570
u can find out the equation by using the mode EQN then unknowns press 3 u enter the number u can get it
so i get x = 2 , y = 1 and z = -1

Answer 4

x+y+z=2
3x-y+z=4(+y & -y cancels)

x+z=2
3x+z=4
(-) (-) (-)
------------
-2x=-2
x=1

from 1 equation
1+z=2
z=1

from equation 3
2*1+y+2*1=3
4+y=3
y= -1

Answer 5

begin with solving

4x + 2z = 6 --> 1st + 2nd eqn.
5x + 3z = 7 --> 2nd + 3rd eqn

find x and z, then find y.

Answer 6

x+y+z=2..................eq(i)
3x-y+z=4................eq(ii)
2x+y+2z=3.............eq(iii)

Adding equation (i) with equation (ii),

x+y+z=2
+(3x-y+z=4)

you get,

4x+2z=6
or, 2z= 6-4x
or, z= 3-2x

Subtracting equation (i) with equation (ii),

x+y+z=2
-(3x-y+z=4)

you get,
-2x+2y= -2
or, x-y=1
or, y=x-1

Then put the value of y and z in eq(iii)

2x+x-1+2(3-2x)=3
or, 2x+x-1+6-4x=3
or, -x=3-5
or, -x =-2
or, x=2

Therefore,
z= 3-2*2
or, z=3-4
or, z= -1

And,
y=2-1
or, y=1

x=2 ; y=1 ; z= -1

Answer 7

this can be sloved by matrix methodassumeit ax=b