Problem 15 - College Algebra

Question

Solve each system by substitution. $$ \begin{aligned} 5 x-2 y+3 z &=4 \\ -4 x+6 y-7 z &=-1 \\ 3 x+2 y-z &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, \frac{1}{2}, 0)\).

Step 1: Solve for One Variable

We'll solve the third equation for \( z \): \[ z = 3x + 2y - 4 \].

Step 2: Substitute for z in the Other Equations

Substitute \( z = 3x + 2y - 4 \) into the first and the second equations: 1. For the first equation:\[ 5x - 2y + 3(3x + 2y - 4) = 4 \] 2. For the second equation:\[ -4x + 6y - 7(3x + 2y - 4) = -1 \]

Step 3: Simplify the Resulting Equations

Simplify and combine like terms in the substituted equations.1. Simplifying the first equation:\[ 5x - 2y + 9x + 6y - 12 = 4 \] Combine like terms:\[ 14x + 4y = 16 \]2. Simplifying the second equation:\[ -4x + 6y - 21x - 14y + 28 = -1 \] Combine like terms:\[ -25x - 8y = -29 \]

Step 4: Solve the New System of Equations for x and y

Solve the system: 1. \( 14x + 4y = 16 \)2. \( -25x - 8y = -29 \)Multiply the first equation by 2 to align \( y \) terms:\[ 28x + 8y = 32 \]Now add it to the second equation to eliminate \( y \):\[ (28x + 8y) + (-25x - 8y) = 32 - 29 \]\[ 3x = 3 \]\[ x = 1 \].

Step 5: Substitute Back to Find y

Substitute \( x = 1 \) into \( 14x + 4y = 16 \):\[ 14(1) + 4y = 16 \]\[ 14 + 4y = 16 \]\[ 4y = 2 \]\[ y = \frac{1}{2} \].

Step 6: Solve for z using x and y

Now substitute \( x = 1 \) and \( y = \frac{1}{2} \) back into the equation for \( z \):\[ z = 3(1) + 2\left(\frac{1}{2}\right) - 4 \]\[ z = 3 + 1 - 4 \]\[ z = 0 \].

Key Concepts

Substitution Method

The substitution method is a way of solving systems of equations by replacing one variable with an equivalent expression containing the other variables. It's like solving a puzzle where you're replacing one piece to see the bigger picture.
First, you need to solve one of the equations for one variable in terms of the others. In our exercise, we solved the third equation for \( z \), which gives us the expression \( z = 3x + 2y - 4 \).
Next, you substitute this expression back into the other equations. This means wherever you see \( z \), you replace it with \( 3x + 2y - 4 \). This helps in reducing the number of variables in the other equations, making them easier to solve.
Finally, you simplify the substituted equations to form a new system of equations that typically involves fewer variables. The key here is to methodically replace, simplify, and solve, step by step.

Linear Equations

Linear equations in a system are usually given in the form \( ax + by + cz = d \), where \( a \), \( b \), and \( c \) are coefficients, and \( d \) is a constant. These equations graphically represent lines in a coordinate space.
The main goal when dealing with linear equations in a system is to find where these lines intersect, which represent the solution to the system. This intersection point gives us the values of the variables that satisfy all the equations simultaneously.
In our exercise, we started with three linear equations. By simplifying one of them in terms of one variable and substituting back, we reduced the problem from three equations in three variables to simpler two-variable forms. Then, these can be addressed using standard algebraic techniques, helping us inch closer to the solution.

Solving for Variables

Solving for variables in a system of linear equations involves finding the specific values that satisfy all the equations at once. First, focus on isolating one variable, as we did by solving for \( z \) in our exercise.
Once you have an equation with a single variable, you substitute this back into another equation, simplifying the terms to combine like variables. For example, plugging the expression for \( z \) into the original equations allowed us to isolate and solve for \( x \) and \( y \) separately.
After substituting, you're left with simpler equations like \( 14x + 4y = 16 \) from the original complex system. Solving these equations, we found \( x = 1 \) and \( y = \frac{1}{2} \).
Finally, with values for \( x \) and \( y \), you determine \( z \) by substituting back into your earlier expression \( z = 3x + 2y - 4 \). The complete solution for the system is thus found as \( x = 1 \), \( y = \frac{1}{2} \), and \( z = 0 \), each fulfilling all original equations.