Problem 42 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{aligned} -\frac{x}{5}+\frac{y}{2} &=-3 \\ \frac{x}{4}-\frac{y}{4} &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution of the system is x = -10 and y = -10.

Step 1: Rearrange one equation to isolate a variable

Let’s take the second equation \( \frac{x}{4}-\frac{y}{4}=0 \), and rearrange it to isolate \( x \): \[ x=y \]

Step 2: Substitute the above value of x into the first equation

Substitute \( x = y \) into the first equation \( -\frac{x}{5}+\frac{y}{2}=-3 \), resulting in : \[ -\frac{y}{5}+\frac{y}{2}=-3 \] then, to solve this equation for \( y \), multiply each term by 10 (the least common multiple of 5 and 2), resulting in: \[ -2y + 5y = -30 \]. This simplifies to: \[ 3y = -30 \]

Step 3: Solve for y

Solve this simplified equation for \( y \) by dividing each side by 3, yielding \( y = -10 \)

Step 4: Substitute y into x=y to solve for x

Now, substitute \( y = -10 \) into the expression \( x = y \) to find the value of \( x \), yielding \( x = -10 \)

Key Concepts

Understanding Systems of Equations

A system of equations involves multiple equations that share common variables. In this case, the system consists of two linear equations. The objective is to find values for each variable that satisfy all equations simultaneously. Variables here might be represented by letters like \( x \) and \( y \), and solving a system means identifying the points where these equations intersect if graphed on a coordinate plane.

There are several methods to solve systems of equations, such as substitution, elimination, or graphing. Solving using substitution, which we focus on here, is particularly helpful when one of the equations can easily be solved for a single variable. This way, you can express one variable in terms of the other, making the process simpler.

Remember, the solution to a system of linear equations can be a single point—where the lines intersect—no solution—if the lines are parallel, or infinitely many solutions—if the lines coincide.

The Process of Solving Linear Equations

Solving a linear equation involves finding the values of the variables that make the equation true. Linear refers to the highest power of the variable being one. It's all about balancing both sides of the equation using operations that maintain equality.

In this exercise, we begin by isolating a variable in one equation. This means getting that variable by itself on one side of the equation to express it in terms of the other. Here, we isolated \( x \) in the second equation, setting it equal to \( y \) (so \( x = y \)). This made it possible to substitute directly into the other equation.

The steps are clear:

• Rearrange one equation to solve for one variable.
• Substitute that variable into the other equation.
• Solve the new equation to find the value of one variable.
• Substitute back to find the second variable.

Mastering Algebraic Manipulation

Algebraic manipulation means performing operations to rearrange equations and simplify expressions. It’s an essential skill that allows you to shift terms across the equal sign, manipulate coefficients, and combine terms.

In our problem, the step of substituting \( x = y \) into the first equation demonstrates this. We then simplified the terms \(- \frac{y}{5} + \frac{y}{2} = -3\) by finding a common denominator or multiplying through to clear fractions. Multiplying each term by 10, a strategic choice here due to it being the least common multiple of 2 and 5, made it easier to proceed.

This simplified our equation to \( -2y + 5y = -30 \). Further combining like terms, \( 3y = -30 \), and then dividing by 3, gave us \( y = -10 \). This is a classic example of simplifying equations using fractions and variables to unlock the solution.