how do you solve the system of equations 3x+4y+z=7; 2y+z=3; -5x+3y+8z=-31 not using matrices?

Answer 1

you are going to have to use SUBSTITUTION. meaning place the equations one above the other. See what you can cross out... i.e.

1) TAKE OUT THE X'S BY MULTIPLYING THE FIRST EQUATION BY 5 AND THE SECOND EQUATION BY 3. YOU WILL GET...

15x+20y+4z=35.
-15x+9y+24z= -93

thus, you can cross out the x's and get the equations:

20y+4z=35
9y+24z= -93, and now you must cross out the z's by multiplying the top equation by -6...

-120y-24z=-210
9y+24z= -93

and, so on....

Answer 2

Step a million: resolve equation (2) for z z = 3 - 2y Step 2: replace consequence of step a million into equation (a million). 3x + 4y + (3 - 2y) = 7 3x + 4y - 2y = 7 -3 3x + 2y = 4 x= (4-2y)/3 Step 3: replace results of step a million and step 2 into equation (3) -5x+3y+8z=-31 -5(4-2y)/3 +3y + 8(3-2y) = -31 -5(4-2y) + 9y + 24(3-2y) = -ninety 3 -20 +10y +9y +seventy two -48y = -ninety 3 10y + 9y -48y = -ninety 3 -seventy two +20 -29y = -one hundred forty five y = -one hundred forty five/-29 = 5 Step 4: use results of step 3 to ensure fee for z utilizing equation (2): 2y + z =3 2(5) + z = 3 10 + z = 3 z = 3 -10 = -7 Step 5: resolve for x by substituting y into answer for step 2 x= (4-2y)/3 x= (4-10)/3 x=-6/3 = -2 There are different approaches to alter as nicely, yet that's one answer.

Answer 3

System of equations:

1) 3x + 4y + z = 7
2) 2y + z = 3
3) -5x + 3y + 8z = -31

From equation 2): z = 3 - 2y

Put the value of z on equation 1) and 3), and you'll obtain:

z in 1):
3x + 4y + (3-2y) = 7
3x + 2y + 3 = 7
3x + 2y =4

z in 2):
-5x + 3y + 8*(3-2y) = -31
-5x + 3y + 24 - 16y = -31
-5x -13y = -55

Now, you have a new system:

1-) 3x + 2y =4
2-) -5x -13y = -55

Multiplying, equation 1-) *5/3:
5x + 10/3 y = 20/3

Adding:
5x + 10/3 y = 20/3
-5x - 13y = -55
-------------------------------------
0 -29/3 y = -145/3
-29 y = -145
y= -145/-29

y=5

Now, we calculate, the value of x, with y=5 in equation 3x + 2y =4
3x + 2(5)=4
3x= 4-10
x=-6/3

x=-2

And we calculate z, with y=5 in equation z = 3 - 2y:
z= 3 - 2*(5)
z= 3 -10

z= -7

Solution: x= -2 ; y =5; z= -7

Answer 4

The way I was taught in school was take the same letter and add or subtract the equation. Once you do that go do it to the rest of the equation and you can figure it out or if its out of a school book look in the back for some of the answers.

Answer 5

3x + 4y + z = 7
2y + z = 3
-5x + 3y + 8z = -31

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

3x + 4y + (-2y + 3) = 7
3x + 4y - 2y + 3 = 7
3x + 2y = 4

-5x + 3y + 8(-2y + 3) = -31
-5x + 3y - 16y + 24 = -31
-5x - 13y = -55

3x + 2y = 4
-5x - 13y = -55

Multiply top by 5 and bottom by 3

15x + 10y = 20
-15x - 39y = -165

-29y = -145
y = 5

z = -2y + 3
z = -2(5) + 3
z = -10 + 3
z = -7

3x + 4y + z = 7
3x + 4(5) - 7 = 7
3x + 20 - 7 = 7
3x + 13 = 7
3x = -6
x = -2

ANS :
x = -2
y = 5
z = -7

Answer 6

Use Solving method of Substitution and Elimination. It is very easy. Do you know how to use these methods. You can take first two equation and solve them in terms of each other.
I hope it helps. For more info Check your chapter on Solving equations.

Answer 7

I hope this helps you ! Below are 3 equation the x1 responds to the coefficient of ''x'' in the first eqation you have. .X3 Is coefficient of x in third eqaution. y1 is coefficient of y in first equation and so forth. Notice the numbers with out a variable (denoted as n1,n2,n3 respectively) must be isolated by its self for each equation, then substitue and then solve for Z.
x1+y1+z1=n1
x2+y2+z2=n2 divison bar over soluiton FO Z=
x3+y3+z3=n3
y3(x1n2-x2n1)+y2(x3n1-x1n3)+y3(x2n3-x3n)
y3(x1z2-x2z1)+y2(x3z1-x1z3)+y1(x2z3-x3z2)

Answer 8

just replace the variables!

-5{7-4[(3-z)/2]-z}+3[(3-z)/2]+8z+31=0

Answer 9

substitute z=3-2y in the other two to get
3x+2y=4 and
5x-13y=-55 then solve.

Answer 10

well my teacher explained all this to me but 3x+4y+z= 3x+4y+z because u just cant add them bcuz of x and y. and the same thing to the rest of the equations