# How do you solve for 2 free variables? Or at least one of them?

So I have a set of 3 equations with 5 variables each; a, b, c, d, and e.

Xf=X1a+X2b+X3c+X4d+X5e

Yf=Y1a+Y2b+Y3c+Y4d+Y5e

Zf=Z1a+Z2b+Z3c+Z4d+Z5e

After I put it through a matrix I'm left with 3 new equations in the form of;

a =Xf+ X4d+X5e

b = Yf+Y4d+Y5e

c = Zf+Z4d+Z5e

(I didn't know how to do subscript)Keep in mind that a, b, c, d, and e are the same as the same variable in each equation. Is there any way to solve for d and e?

## 1 Answer

Suppose we have

a =Xf+ X4d+X5e

b = Yf+Y4d+Y5e

c = Zf+Z4d+Z5e

Then

X4d+X5e=a -Xf

Y4d+Y5e =b -Yf

Z4d+Z5e= c -Zf

This is a linear system of equations with 2 unknowns and 3 equations. In general, such equation does not have a solution, unless the equations are not linearly independent, i.e., one of the equations is a linear combination of the other two.

- 1 Answer
- 162 views
- Pro Bono

### Related Questions

- The Span and Uniqueness of Solutions in a Parametric Matrix
- Consider the matrix, calculate a basis of the null space and column space
- Find the null space of the matrix $\begin{pmatrix} 1 & 2 & -1 \\ 3 & -3 & 1 \end{pmatrix}$
- Linear Algebra Assistance: Linear Combinations of Vectors
- Linear Algebra Question
- [change of basis] Consider the family β = (1 + x + x 2 , x − x 2 , 2 + x 2 ) of the polynomial space of degree ≤ 2, R2[x].
- Find the general solution of the system of ODE $X'=\begin{bmatrix} 1 & 3 \\ -3 & 1 \end{bmatrix} X$
- Singular Value Decomposition Example

Your question does not make sense to me. What do you mean by "After I put it through a matrix"? In general, in order to solve linear equations the number of variables and equations should be equal.

To reduce the first set of equations I put it into a 3x6 matrix. I'm aware that leaves d and e as free variables, I'm wondering if there is a way to solve for them.