Problem 33 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} &x+y+z=83,\\\ &y=2 x+3,\\\ &z=40+x \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 10 \), \( y = 23 \), \( z = 50 \).

Step 1: Substitute Equation for y

Use the equation for y provided. Substitute the expression for y from the second equation into the first equation. \[ y = 2x + 3 \] Substitute this into the first equation: \[ x + (2x + 3) + z = 83 \]

Step 2: Substitute Equation for z

Next, use the third equation for z. Substitute the expression for z from the third equation into the modified first equation. \[ z = 40 + x \] Substitute this into the equation: \[ x + (2x + 3) + (40 + x) = 83 \]

Step 3: Simplify and Solve for x

Combine like terms and solve for x: \[ x + 2x + 3 + 40 + x = 83 \] \[ 4x + 43 = 83 \] Subtract 43 from both sides: \[ 4x = 40 \] Divide by 4: \[ x = 10 \]

Step 4: Solve for y

Substitute the value of x back into the equation for y: \[ y = 2x + 3 \] \[ y = 2(10) + 3 = 20 + 3 = 23 \]

Step 5: Solve for z

Substitute the value of x back into the equation for z: \[ z = 40 + x \] \[ z = 40 + 10 = 50 \]

Step 6: Verify the Solution

Check that the values satisfy the original equations. First equation: \[ x + y + z = 10 + 23 + 50 = 83 \] Second equation: \[ y = 2x + 3 = 2(10) + 3 = 23 \] Third equation: \[ z = 40 + x = 40 + 10 = 50 \] All equations are satisfied, so the solution is correct.

Key Concepts

simultaneous equations

Simultaneous equations, also known as systems of equations, are a set of equations containing multiple variables. The goal is to find values for each variable that will satisfy all equations in the system at the same time.

In our given problem, we have three equations with three variables: x, y, and z.

• The first equation is: \( x + y + z = 83 \)
• The second equation is: \( y = 2x + 3 \)
• The third equation is: \( z = 40 + x \)

Each of these equations represents a relationship between the variables. When we solve these equations simultaneously, we're looking for one set of variable values that makes all three equations true at the same time.

substitution method

The substitution method is a way to solve a system of equations by solving one of the equations for one variable and then substituting that expression into the other equations.

Let's break it down using our problem:

• First, notice the second equation tells us that \( y = 2x + 3 \).
• We'll substitute this into the first equation \(x + (2x + 3) + z = 83\).
• This gives us \( x + 2x + 3 + z = 83 \).

Now, let's use the third equation:

• We see that \( z = 40 + x \).
• Substitute \(40 + x \) for z in our modified first equation: \( x + 2x + 3 + (40 + x) = 83 \).
• This becomes \( 4x + 43 = 83 \).
• Subtract 43 from both sides: \( 4x = 40 \).
• Finally, divide by 4 to find \( x = 10 \).

The substitution method helps simplify the solution process by reducing the number of equations and making it easier to solve for each variable.

linear equations

Linear equations are equations where the highest power of the variable(s) is 1. These equations graph as straight lines on the Cartesian plane.

In our problem, we deal with linear equations:

• \( x + y + z = 83 \)
• \( y = 2x + 3 \)
• \( z = 40 + x \)

Each equation can be simplified and solved using basic algebraic operations. Linear equations are fundamental in algebra because they are straightforward to manipulate and solve. Let's look at how we used them in our steps:

First, we isolated variables using the simpler equations (steps 1 and 2).

We substituted these expressions into the more complex equations.

Finally, we combined like terms and solved for our primary variable. This systematic approach helps ensure we find the correct values for all variables in the system.

Once we solve for one variable, the solutions for the remaining variables fall into place through substitution back into the original equations. This robustness of linear equations makes them a staple in problem-solving.
By thoroughly understanding linear equations, simultaneous equations, and the substitution method, you can confidently tackle a wide variety of algebraic problems!