Problem 51 - Elementary Algebra Within Reach

Question

In Exercises 50-53, solve the system by the method of elimination. $$ \left\\{\begin{array}{l} -4 x+3 y=18 \\ -6 x+y=-8 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = 3 \), \( y = 10 \)

Step 1: Multiply the Equations by Suitable Numbers to make the Coefficient of y's or x's Same

Multiply the first equation by 1 and the second equation by 3. The resulting system of equations is: \[ \begin{cases} -4x + 3y = 18 \ -18x + 3y = -24 \end{cases} \]

Step 2: Subtract the Two Equations to Eliminate y

Subtract the second equation from the first equation which will result in: \[ -4x -(-18x) = 18 -(-24) \], which simplifies to \( 14x = 42 \)

Step 3: Solve for the Remaining Variable x

Solve for x in the equation \(14x = 42\) to get \( x = 3 \)

Step 4: Substitute the Value of x into one of the Original Equations to Solve for y

Substitute \( x = 3 \) into the original equation \( -4x + 3y = 18 \) to get: \[ -4(3) + 3y = 18 \] which simplifies to \( 3y = 30 \), upon solving this we get \( y = 10 \)

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations that share the same set of unknown variables. In this particular exercise, we have a system of two equations involving two variables, \( x \) and \( y \). These equations are:

• \(-4x + 3y = 18\)
• \(-6x + y = -8\)

The goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously. Such systems can be solved using various methods, but the elimination method is often a very effective strategy when the coefficients can be easily manipulated. By eliminating one variable, we simplify the problem, allowing us to solve for the remaining variable. Once one has a value, it is substituted back to find the other variable.

Linear Equations

Linear equations are equations where the highest power of any variable is one. Each term in a linear equation can be a constant or a product of a constant and a single variable. In this exercise, each of the given equations is linear, hence they graph as straight lines in the coordinate plane. An equation like \(-4x + 3y = 18\) fits this category perfectly.Linear equations can be visualized graphically where each equation is represented as a line on a coordinate plane. The point where these lines intersect is the solution of the system. This intersection represents the values of \(x\) and \(y\) that satisfy both equations. The elimination method simplifies this by removing one variable, allowing for easier algebraic manipulation to find where these lines meet.

Algebraic Solutions

Algebraic solutions involve manipulating equations to find the variable values that satisfy all equations in a given system. Using algebra, we can accurately calculate precise solutions.In the elimination method, we start by making the coefficients of one of the variables equal in both equations, as was done by multiplying the equations by appropriate numbers. In this case, multiplying the second equation by \(3\) aligns the coefficients of \( y \).By subtracting the resulting equations, we eliminate \( y \) and solve for \( x \). Solving \( 14x = 42 \) gives \( x = 3 \). This value is plugged back into one of the original equations to solve for \( y \), resulting in \( y = 10 \). Through disciplined algebraic manipulation, solutions are found efficiently.