# Help Solving Systems of Equations in 3 Variables

• Mar 26th 2007, 05:54 PM
lilrhino
Help Solving Systems of Equations in 3 Variables
Hi All,

My issue is solving 3 systems of variables that are missing a variable in each equation.

My 3 systems are:

3p + 2r = 11, (1)
q - 7r = 4, (2)
p - q = 1 (3)

What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

With that said, I proceeded with:

Multiplied equation (2) by 1 and added it to equation (3)

q - 7r = 4
p - q =1
----------
p - 7r = 5 (4)

Next, I multiplied equation (4) by - 3 and added it to equation (1)

3p + 2r = 11
-3p + 21r = -15
----------------
23r = -4
r = -4/23 (fraction)

Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...
• Mar 26th 2007, 06:14 PM
topsquark
Quote:

Originally Posted by lilrhino
Hi All,

My issue is solving 3 systems of variables that are missing a variable in each equation.

My 3 systems are:

3p + 2r = 11, (1)
q - 7r = 4, (2)
p - q = 1 (3)

What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

With that said, I proceeded with:

Multiplied equation (2) by 1 and added it to equation (3)

q - 7r = 4
p - q =1
----------
p - 7r = 5 (4)

Next, I multiplied equation (4) by - 3 and added it to equation (1)

3p + 2r = 11
-3p + 21r = -15
----------------
23r = -4
r = -4/23 (fraction)

Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...

You are doing great! Don't stop now on account of a few fractions:

r = -4/23

p - 7r = 5 ==> p - 7(-4/23) = 5

p + 28/23 = 5

p = 5 - 28/23

p = 115/23 - 28/23 = 87/23

And
p - q = 1 ==> 87/23 - q = 1

q = 87/23 - 1 = 87/23 - 23/23 = 64/23

p = 87/23
q = 64/23
r = -4/23

-Dan
• Mar 26th 2007, 06:17 PM
lilrhino
Quote:

Originally Posted by topsquark
You are doing great! Don't stop now on account of a few fractions:

r = -4/23

p - 7r = 5 ==> p - 7(-4/23) = 5

p + 28/23 = 5

p = 5 - 28/23

p = 115/23 - 28/23 = 87/23

And
p - q = 1 ==> 87/23 - q = 1

q = 87/23 - 1 = 87/23 - 23/23 = 64/23

p = 87/23
q = 64/23
r = -4/23

-Dan

Hi Dan, thanks...I got q = 64/23, but I stopped solving because I thought it was incorrect. Thanks a bunch! :)
• Oct 12th 2007, 02:39 AM
rich.sisk
help me trying to so problem for a variable
how do i solve 3x=wyz i need to solve
for the variable x
for the variable w
for the variable y
for the variable z
plz need help trying to cheach up in class so i can pass any one is welcome

Quote:

Originally Posted by lilrhino
Hi All,

My issue is solving 3 systems of variables that are missing a variable in each equation.

My 3 systems are:

3p + 2r = 11, (1)
q - 7r = 4, (2)
p - q = 1 (3)

What I'm seeing is - that there's no q in equation (1). That means that q has been eliminated from one equation. Since I need another equation with q eliminated, I wanted to use equations (2) and (3) to eliminate q.

With that said, I proceeded with:

Multiplied equation (2) by 1 and added it to equation (3)

q - 7r = 4
p - q =1
----------
p - 7r = 5 (4)

Next, I multiplied equation (4) by - 3 and added it to equation (1)

3p + 2r = 11
-3p + 21r = -15
----------------
23r = -4
r = -4/23 (fraction)

Whenever, I get fraction that confuses me as to how to solve for the other variables. I plugged -4/23 back into equation (1) to solve for p and got an improper fraction. Is this correct? Or, can someone provide some assistance with how I should've proceeded in the first place? I'm missing something here and I'm not sure how I should proceed? Please help. TIA...

• Oct 12th 2007, 04:32 AM
earboth
Quote:

Originally Posted by rich.sisk
how do i solve 3x=wyz i need to solve
for the variable x
for the variable w
for the variable y
for the variable z
plz need help trying to cheach up in class so i can pass any one is welcome

Hello,

one remark at first: If you have a new question please start a new thread. Otherwise you risk that nobody will notice that you need some help.

Solving for a variable means that you collect all terms containing this variable at one side of the equation and all other terms on the other side. Presumely you have to get rid of some coefficients to isolate the variable:

$\displaystyle 3x=wyz~\implies~\text{ divide by 3: }~\frac{3x}{3} =x= \frac13 \cdot w \cdot y \cdot z$

$\displaystyle 3x=wyz~\implies~\text{ divide by wz: }~y = \frac{3x}{ w \cdot z}$

$\displaystyle 3x=wyz~\implies~\text{ divide by yz: }~w = \frac{3x}{ y \cdot z}$

$\displaystyle 3x=wyz~\implies~\text{ divide by wy: }~z = \frac{3x}{ w \cdot y}$