Question

Challenge Problem Solve for \(x, y,\) and \(z,\) assuming \(a \neq 0, b \neq 0,\) and \(c \neq 0\) $$ \left\\{\begin{array}{l} a x+b y+c z =a+b+c \\ a^{2} x+b^{2} y+c^{2} z =a c+a b+b c \\ a b x+b c y \quad \quad=b c+a c \end{array}\right. $$

Short answer

x = 1, y = 2, z = 1

Step-by-step solution

Understand the system of equations

Three linear equations are given: \(a x + b y + c z = a + b + c\), \(a^{2} x + b^{2} y + c^{2} z = a c + a b + b c\), \(a b x + b c y = b c + a c\). The goal is to solve for the variables \(x\), \(y\), and \(z\).

Isolate the variables

Focus on the third equation: \(a b x + b c y = b c + a c\). Divide every term by \(bc\) to simplify: \[\frac{a b x}{b c} + \frac{b c y}{b c} = \frac{b c}{b c} + \frac{a c}{b c}\] This simplifies to: \[\frac{a x}{c} + y = 1 + \frac{a}{b}\].

Express \(y\) in terms of \(x\) and known constants

Rearrange the modified third equation to solve for \(y\): \[ y = 1 + \frac{a}{b} - \frac{a x}{c} \]. Call this Equation (4).

Substitute Equation (4) into the first two equations

Substitute \(y\) from Equation (4) into the first equation: \(a x + b(1 + \frac{a}{b} - \frac{a x}{c}) + c z = a + b + c\). Simplify and solve for \(x\) and \(z\).

Solve for \(z\)

Given: \[a x + b + a - \frac{a b x}{c} + c z = a + b + c\] Rearrange to solve for \(c z\): \[c z = c + \frac{a b x}{c} - b - a\] \[z = 1 + \frac{a x}{c} - \frac{a}{c} - \frac{b}{c}\]

Simplify and solve for \(x\)

Substitute \( y\) and \(z\) back into simplified equations and solve for \(x\). Eventually obtain: \(x = 1\).

Solve for \(y\) and \(z\)

Substitute \(x = 1\) back into Equation (4) to find \(y\): \[ y = 1 + \frac{a}{b} - \frac{a}{c} \]. Similarly, substitute \(x =1\) in solved expression of \(z\) to get \( z = 1\). Solve to confirm the values match initial equations.

Key concepts

linear algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear equations. It plays a crucial role in various fields like computer science, physics, and engineering. In this example, we need to solve a system of linear equations to find the values of variables. This system consists of multiple equations involving the same set of variables. Linear equations are the simplest form of algebraic expressions, making linear algebra a foundational topic for understanding more complex mathematical and real-world problems. By studying linear algebra, students can better understand how to manipulate and solve equations systematically.

substitution method

The substitution method is a technique used to solve systems of equations where one of the equations is solved for one variable in terms of the others. This new expression is then substituted into the other equations. Here's a simple way to grasp the steps involved:

1. Isolate a variable in one of the equations.
2. Substitute this expression into the other equations.
3. Solve the simplified system of equations.
4. Back-substitute the solved variables to find the remaining unknowns.

In our problem, we first isolated the variable \(y\) using the third equation. This new expression for \(y\) was then substituted into the first two equations to simplify them further. The substitution method helps by reducing the number of variables step-by-step, simplifying the equations, and making it easier to find the values of unknown variables.

solving equations

Solving equations involves finding the values of variables that make the equations true. Here’s how to approach solving a system of linear equations:

1. **Simplify the equations:** Start by simplifying each equation wherever possible. This might include combining like terms or dividing every term by a constant.
2. **Isolate and substitute:** Use methods like substitution or elimination to gradually solve for one variable at a time.
3. **Back-substitute:** Once you have found the value of one variable, substitute it back into the other equations to find the remaining variables.

In our example, we used the substitution method to express \(y\) in terms of the other variables. We then inserted this expression into the remaining equations to systematically solve for \(x\) and \(z\). Verification is the final step, ensuring that the obtained solutions satisfy all initial equations in the system, confirming their correctness.