Jay rewrites a system of 2 linear equations as the augmented matrix below:
 
	–4	9	16	
10	3	62

Jay then uses a matrix calculator to convert the matrix to reduced row echelon form, allowing her to see the solutions to x and y.



What is the value of x in the solution to Jay’s system

We are given the augmented matrix:

\[
\begin{pmatrix}
-4 & 9 & 16 \\
10 & 3 & 62
\end{pmatrix}
\]

To solve for \(x\) and \(y\), we need to convert this matrix into reduced row echelon form (RREF). The RREF will provide us with the solutions to \(x\) and \(y\).

### Step 1: Row Operations to Achieve RREF
We want to make the first entry in the top left of the matrix \(1\), and the rest of the entries below it and in the same column \(0\), followed by similar operations to get a leading 1 in the second row.

Let's outline the necessary row operations.

#### Current Matrix:
\[
\begin{pmatrix}
-4 & 9 & 16 \\
10 & 3 & 62
\end{pmatrix}
\]

#### Row 1 (R1): Divide R1 by \(-4\) to get a leading 1 in the first position:
\[
\begin{pmatrix}
1 & -\frac{9}{4} & -4 \\
10 & 3 & 62
\end{pmatrix}
\]

#### Row 2 (R2): Subtract \(10 \times \text{R1}\) from R2 to eliminate the \(10\) below the first pivot:
\[
\begin{pmatrix}
1 & -\frac{9}{4} & -4 \\
0 & \frac{93}{4} & 102
\end{pmatrix}
\]

#### Row 2: Divide R2 by \(\frac{93}{4}\) to make the pivot in R2 a 1:
\[
\begin{pmatrix}
1 & -\frac{9}{4} & -4 \\
0 & 1 & \frac{408}{93}
\end{pmatrix}
\]
(Simplifying \(\frac{408}{93}\) gives \(\frac{136}{31}\)).

#### Row 1: Add \(\frac{9}{4} \times \text{R2}\) to R1 to eliminate the \(-\frac{9}{4}\) in the first row:
\[
\begin{pmatrix}
1 & 0 & -\frac{80}{31} \\
0 & 1 & \frac{136}{31}
\end{pmatrix}
\]

### Final Solution
The reduced row echelon form gives us the system:
\[
x = -\frac{80}{31}, \quad y = \frac{136}{31}
\]

Thus, the value of \(x\) is:

\[
x = -\frac{80}{31}
\]

Would you like a detailed breakdown of any step, or do you have questions on how RREF works?

Here are five related questions for further exploration:
1. How can you determine the value of \(y\) from the matrix?
2. What are some key row operations used to convert matrices into RREF?
3. How does the augmented matrix relate to the original system of linear equations?
4. What is the significance of the leading 1's in the RREF?
5. Can you solve a system of three linear equations using the same process?

**Tip:** When using RREF, ensure the leading coefficient (pivot) in each row is 1, and all other entries in that column are zero.

Jay rewrites a system of 2 linear equations as the augmented matrix below:

-4 9 16
10 3 62

Jay then uses a matrix calculator to convert the matrix to reduced row echelon form, allowing her to see the solutions to x and y. What is the value of x in the solution to Jay’s system?

Learn how to solve a system of linear equations by converting an augmented matrix into reduced row echelon form (RREF). This problem involves matrix operations and provides a step-by-step solution to find the value of x for Grades 10-12 students.

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