Problem 49 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. $$ \begin{array}{l}{6 x+y=15} \\ {x-4 y=-10}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \) and \( y = 3 \).

Step 1: Choose the method of substitution

First, we choose substitution or elimination. In this case, we will use substitution as one of the equations can be easily solved for one variable. Let's solve the second equation for \( x \).

Step 2: Solve for x in the second equation

Take the equation \( x - 4y = -10 \) and solve for \( x \). Add \( 4y \) to both sides to get \( x = 4y - 10 \).

Step 3: Substitute x in the first equation

Substitute \( x = 4y - 10 \) into the first equation \( 6x + y = 15 \). This gives \( 6(4y - 10) + y = 15 \).

Step 4: Solve for y

Distribute and simplify \( 6(4y - 10) + y = 15 \) to get \( 24y - 60 + y = 15 \). Combine like terms to obtain \( 25y - 60 = 15 \). Add 60 to both sides, giving \( 25y = 75 \). Divide both sides by 25 to find \( y = 3 \).

Step 5: Substitute y back to find x

Now that we know \( y = 3 \), substitute this value back into \( x = 4y - 10 \). This becomes \( x = 4(3) - 10 = 12 - 10 = 2 \).

Step 6: Verify the solution

Plug \( x = 2 \) and \( y = 3 \) back into the original equations to verify. For \( 6x + y = 15 \): \( 6(2) + 3 = 12 + 3 = 15\), which is correct. For \( x - 4y = -10 \): \( 2 - 4(3) = 2 - 12 = -10 \), which is also correct.

Key Concepts

Substitution Method

The substitution method is a way to solve systems of equations by solving one of the equations for one variable in terms of the others. In this exercise, we chose to use substitution because one equation can be easily solved for a single variable. Here’s how you can think about it:

• Select one equation to solve for one of the variables. This depends on which equation and which variable is the simplest to isolate.
• Solve the selected equation for the chosen variable. This involves using basic algebraic manipulations such as adding, subtracting, multiplying, or dividing both sides of the equation to isolate the variable.
• Once you have solved for one variable, substitute this expression into the other equation. This reduces the system to a single equation with one variable.

In our problem, we solved for \( x \) in the second equation because it was straightforward to do so. This gave us \( x = 4y - 10 \). The expression \( x = 4y - 10 \) was then substituted into the first equation, simplifying our process considerably.

Elimination Method

While the substitution method was used in this problem, the elimination method is another powerful technique for solving systems of equations. It involves adding or subtracting the equations to eliminate one variable, making it easier to solve for the other. This method is particularly useful when both equations are in standard form and arranged such that one variable can be easily cancelled out. Here’s a quick overview:

• Align the equations, making sure both are in standard form.
• If necessary, multiply one or both equations by a constant so that adding or subtracting them will eliminate one variable.
• Add or subtract the equations to eliminate a variable, yielding a single equation with one variable remaining.

In our original system, an alternative approach could have involved multiplying the second equation by a number that would allow us to add it to or subtract it from the first equation, thus cancelling out one of the variables. This flexibility makes elimination a handy method in many situations.

Algebraic Solution Steps

Regardless of the method, solving systems of equations requires a clear understanding of algebraic steps. Let's review these steps using the context of solving our specific problem.Firstly, identify what you are solving for: in a system of equations, it's two variables. Choose the method based on simplicity or preference, such as substitution for easily isolated variables or elimination for clear equation alignment.In the algebra of our problem:

• We first solved the second equation \( x - 4y = -10 \) to find \( x \) as \( 4y - 10 \).
• Then we substituted \( x = 4y - 10 \) into the first equation \( 6x + y = 15 \), which transformed it into a single-variable equation \( 24y - 60 + y = 15 \).
• After simplifying and solving \( 25y = 75 \), we found \( y = 3 \).
• Substituting back, \( x = 4(3) - 10 \), the solution was verified as \( x = 2 \).

Every step uses basic algebra: combining like terms, isolating variables, and verifying solutions, emphasizing the importance of both simplifying and checking your work.