91Ó°ÊÓ

A system of equations is a collection of two or more equations that share common variables. In mathematical terms, the given calculation features three equations with variables \(x\), \(y\), and \(z\). Solving a system of equations involves finding a set of values for these variables that satisfy all the equations simultaneously.
In our exercise, the system is represented as:

• \(x + 5y - 3z = -2\)
• \(-3x - 16y + 7z = -\frac{1}{2}\)
• \(-x - 5y + 4z = 0\)

These equations are interconnected, and each one provides essential information about the relationships between \(x\), \(y\), and \(z\). The goal is to find the values of these variables that make all three equations correct simultaneously. Success in finding the solution means the values work in each of the original equations. Solving systems of equations is foundational in algebra and has applications in various fields such as engineering, physics, and economics.

The substitution method is a technique used to solve systems of equations where one equation is solved for one variable, and then substituted into the other equations. This method is especially effective when one of the equations is easy to solve for a single variable.
In our solution, we first isolate \(x\) in the third equation:

• \(-x - 5y + 4z = 0\) becomes \(x = 5y - 4z\)

By solving for \(x\), we can express it in terms of \(y\) and \(z\). The key advantage here is simplifying the problem by reducing the number of variables in the other equations. Next, we substitute \(x = 5y - 4z\) into the first two equations. This substitution turns them into two equations with two variables:

• \(10y - 7z = -2\)
• \(-31y + 19z = -0.5\)

The substitution method essentially transforms a complex problem into a simpler one where it becomes easier to find the remaining variable values.

The elimination method, also known as the addition or subtraction method, involves removing a variable from a system of equations to solve for the remaining variables. This is done by adding or subtracting the equations in such a way that one variable cancels out.
In our exercise, after applying the substitution method, we ended up with two simpler equations:

• \(10y - 7z = -2\)
• \(-31y + 19z = -0.5\)

To solve these efficiently, we aim to eliminate one of the variables. We can do this by:

• Multiplying the first equation by \(31\), resulting in \(310y - 217z = -62\)
• Multiplying the second equation by \(10\), resulting in \(-310y + 190z = -5\)
• Adding the resulting equations to eliminate \(y\): \(310y - 217z + (-310y + 190z) = -62 - 5\)
• This simplifies to \(z = -1\)

Now we substitute \(z = -1\) back into one of the equations to find \(y\). As shown in the steps, back-substitution gives \(y = -0.9\). Finally, plug these values back into the equation for \(x\), and we find \(x = -1\). This comprehensive use of both substitution and elimination provides a robust method for solving complex systems of equations.