# Solving x+3y=4-kz & 4x-2y=5+10z: Find the Solution Here

• chwala
In summary, the conversation discusses finding the solution to a system of linear equations using different methods such as row reduction and determinant. The final solution is found to be k = 8, but it is noted that there are multiple approaches to solving the problem and the wording of the question is important in understanding whether there is a unique solution or not. It is also mentioned that following instructions and using methods precisely is important in mathematical problem solving.
chwala
Gold Member
Homework Statement
See attached
Relevant Equations
linear equations

Find the solution here;

I will need to check which approach they used...probably some row- reduced echelon somewhere... anyway my approach;

##x+3y=4-kz##
##4x-2y=5+10z##
.................................
##4x+12y=16-4kz##
##4x-2y=5+10z##
.................................
##14y=11-4kz-10z##
.................................
From third equation,
##x+y+2z=1##
##y=1-2z-x##
##y=1-2z-(4-kz-3y)##
##-2y=-3-2z+kz## multiplying this by ##7##
.................................
Therefore,
##14y=11-4kz-10z##
##-14y=-21-14z+7zk##
...................................
subtracting we get,
##10=3kz-24z##
##10=z(3k-24)##
.................................
For no unique solution, we set
##3k-24=0##
##k=8##

Probably you may inform me which approach they used...just the form...otherwise, i can check this up later...

Arrrrgh i have seen it...using determinant! cheers guys.

Delta2
Using determinant, we shall have;
##1(-4+10)-3(8+10)+k(4+2)=0##
##6-54+6k=0##
##6k=48##
##k=8## Bingo! learned this many years ago...going deep into my brain to remember...

Beware of the wording in this problem. Comparing the wording in your title ”has no solution” it has quite different meaning from the problem’s ”has no unique solution”.

In this case, there indeed are solutions. They are just not unique.

Orodruin said:
Beware of the wording in this problem. Comparing the wording in your title ”has no solution” it has quite different meaning from the problem’s ”has no unique solution”.

In this case, there indeed are solutions. They are just not unique.
Thanks...let me amend that...

We can also use the row-reduction method,

##\begin{bmatrix}
1 & 3 & k \\
2 & -1 & -5 \\
1 & 1 & 2
\end{bmatrix}##

##2R_1 -R_2##

##\begin{bmatrix}
1 & 3 & k \\
0 & 7 & 2k+5 \\
1 & 1 & 2
\end{bmatrix}##

##R_1-R_3##

##\begin{bmatrix}
1 & 3 & k \\
0 & 7 & 2k+5 \\
0 & 2 & k-2
\end{bmatrix}##

##2R_2-7R_3##

##\begin{bmatrix}
1 & 3 & k \\
0 & 14 & 4k+10 \\
0 & 0 & 7k-14
\end{bmatrix}##

##4k+10-(7k-14)=0##
##-3k+24=0##
##k=8##

Bingo!

Last edited:
Sorry allow me to ask...since ms are allowed to be followed 'strictly' to the latter...does it mean that any other approach/ method to solution wouldn't be accepted? like method shown in post ##1## and ##5.## Cheers.

chwala said:
.since ms are allowed to be followed 'strictly' to the latter
"ms are allowed to be followed... " -- What are "ms"?
Also, the phrase is "to the letter."
chwala said:
.does it mean that any other approach/ method to solution wouldn't be accepted? like method shown in post 1 and 5. Cheers.
Method in post 1 is an approach that would be taken by someone with no knowledge of matrix reduction methods. Method in post 5 relies solely on the determinant to be able to discern whether the matrix of coefficients is invertible.

An alternate method that is better than that used in post 5 is to use row reduction on the augmented matrix. That is, the 3 x 4 matrix that contains a 4th column that has the constants on the right sides of the given three equations.

When I worked the problem, the third row was ##\begin{bmatrix}0 & 0 & -24 + 3k & | &0 \end{bmatrix}##. This means that if k = 8, the bottom row consists solely of 0 elements, which further means that the three given equations are linearly dependent. Graphically, if k = 8, the three planes intersect at the least in a line in space.

On the other hand, if the bottom row had ended up with a nonzero constant in the 4th column, for example like this... ##\begin{bmatrix}0 & 0 & -24 + 3k & | &1 \end{bmatrix}##, that would represent the equation 0x + 0y + 0z = 1 (again with k = 8), which has no solution. In this case the system of equations would have no solution at all; i.e., no single point lies on all three of the given planes.

chwala, Delta2 and Orodruin

## 1. What is the meaning of "solving x+3y=4-kz & 4x-2y=5+10z"?

The given expression represents a system of linear equations, where x, y, and z are variables and the coefficients (numbers multiplied by the variables) are known values. The goal is to find the values of x, y, and z that satisfy both equations simultaneously.

## 2. How do you solve a system of linear equations?

There are several methods for solving a system of linear equations, such as substitution, elimination, and graphing. In this case, the most efficient method would be to use Gaussian elimination, which involves manipulating the equations to eliminate one variable at a time and solve for the remaining variables.

## 3. What does "find the solution" mean in this context?

The solution to a system of linear equations is a set of values for the variables that make both equations true. In other words, it is the point of intersection between the two lines represented by the equations on a graph.

## 4. Is there always a solution to a system of linear equations?

No, there may not always be a solution to a system of linear equations. If the equations are parallel, they will never intersect and there will be no solution. If the equations are identical, there will be infinitely many solutions.

## 5. What is the importance of solving systems of linear equations in science?

Solving systems of linear equations is important in science because many real-world problems can be modeled using linear equations. By finding the solution to these equations, scientists can make predictions and analyze data to better understand and solve complex problems in various fields such as physics, chemistry, and economics.

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