if 23+28+37+x+53=168 and 23+28+40+y+50=120. Find the value of x-y

asked by guest
on Jan 26, 2025 at 9:28 pm



You asked:

Given the following equations: \[\begin{aligned}23 + 28 + 37 + x + 53 &= 168\\ 23 + 28 + 40 + y + 50 &= 120\end{aligned}\] Evaluate \(x - y\).

MathBot Answer:

The system of linear equations has exactly one solution.

The substitutions are \( x = 27 \), \( y = -21 \).

\[x - y = 48\]


Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}1 & 0 &\bigm |& 27\\0 & 1 &\bigm |& -21\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.


First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}1 & 0 &\bigm |& 27\\0 & 1 &\bigm |& -21\end{bmatrix}$$

The leading term of row \(1\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 &\bigm |& 27\\0 & 1 &\bigm |& -21\end{bmatrix}$$

The leading term of row \(2\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 &\bigm |& 27\\0 & 1 &\bigm |& -21\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y = 27 \\ x = 27\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y = -21 \\ y = -21\end{aligned}$$

By substituting these values into the expression, we find that:$$\begin{aligned} x - y &= 27 - -21\\ &=48\end{aligned}$$