Problem 55 - Precalculus with Limits

Question

In Exercises \(55-58\) , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables \(x, y,\) and \(z,\) if applicable.) $$\left[\begin{array}{rrrr}{1} & {-2} & {\vdots} & {4} \\ {0} & {1} & {\vdots} & {-3}\end{array}\right]$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -2, y = -3\).

Step 1: Transform the Augmented Matrix into a System of Linear Equations

The augmented matrix is essentially a compact representation of system of linear equations. Each row corresponds to an equation, and each column represents a variable. In this case, the given matrix represents two equations with two variables. The first row corresponds to the equation \(1x - 2y = 4\) and the second row corresponds to the equation \(0x + 1y = -3\). So, we have two equations: \(x = 2y + 4\) and \(y = -3\).

Step 2: Back-Substitution

Back substitution involves substituting the solutions for y into the equations involving x to find the solutions for x. As y = -3 is already isolated, we substitute y = -3 in the first equation: \(x = 2*(-3) + 4 => x = -2\). Therefore, the solution to the system of equations is \(x = -2, y = -3\).