Problem 34 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}5 x-3 y=-34 \\ 2 x+7 y=-30\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \((x, y) = (-8, -2)\).

Step 1: Solve one equation for one variable

Choose one of the equations to express one variable in terms of the other. Let's solve the first equation for \(x\):\[5x - 3y = -34\]Add \(3y\) to both sides:\[5x = 3y - 34\]Then divide by 5:\[x = \frac{3y - 34}{5}\]

Step 2: Substitute the expression into the other equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation:\[2x + 7y = -30\]Substitute \(x = \frac{3y - 34}{5}\):\[2\left(\frac{3y - 34}{5}\right) + 7y = -30\]

Step 3: Simplify the equation

Multiply through by 5 to eliminate the fraction:\[2(3y - 34) + 35y = -150\]Simplify:\[6y - 68 + 35y = -150\]Combine like terms:\[41y - 68 = -150\]

Step 4: Solve for \(y\)

Add 68 to both sides:\[41y = -82\]Divide by 41:\[y = -2\]

Step 5: Substitute \(y\) back to find \(x\)

Now that we have \(y = -2\), substitute it back into the equation found in Step 1:\[x = \frac{3(-2) - 34}{5}\]Calculate:\[x = \frac{-6 - 34}{5} = \frac{-40}{5} = -8\]

Step 6: State the solution

The solution to the system of equations is \((x, y) = (-8, -2)\). Both equations hold true when \(x = -8\) and \(y = -2\).

Key Concepts

Understanding Systems of Equations

Systems of equations consist of two or more equations that share common variables. The solution to a system of equations is the value for each variable that satisfies all equations simultaneously. In this case, we are dealing with a system consisting of two linear equations:

• Equation 1: \(5x - 3y = -34\)
• Equation 2: \(2x + 7y = -30\)

Understanding that these equations represent lines on a graph helps visualize the solution. The solution is where the lines intersect, meaning both equations are true for those specific values of \(x\) and \(y\). Using algebraic methods like substitution, elimination, or graphical analysis leads us to this intersecting point, also known as the 'solution of the system.'
Substitution is particularly useful when one of the equations is easily solvable for one of the variables.

Solving Linear Equations

Solving linear equations involves finding the value of a variable that makes the equation true. In a single linear equation, you have a straight line on the graph. Let's consider how the substitution method works with the first equation in our system:

• Start by isolating a variable in one equation: \(5x - 3y = -34\) can be rearranged to express \(x\) in terms of \(y\):
• Add \(3y\) to both sides to isolate terms with \(x\): \(5x = 3y - 34\).
• Divide by 5 to get \(x\): \(x = \frac{3y - 34}{5}\).

Now that we have \(x\) in terms of \(y\), this expression can be substituted into the second equation. Solving linear equations usually involves a few steps of algebraic manipulation to isolate variables or simplify expressions, as shown here.

Algebraic Manipulation Techniques

Algebraic manipulation is crucial in solving systems of equations and involves rearranging equations to isolate variables or simplify expressions.
In the substitution method, once we have an expression for one variable, we replace it in the other equation. This step ensures we solve for one variable at a time. Here are the key manipulation steps in our context:

• Substituting \(x = \frac{3y - 34}{5}\) into \(2x + 7y = -30\) results in an equation with a single variable \(y\).
• Eliminate fractions by multiplying the entire equation by 5, simplifying the process: \(2(3y - 34) + 35y = -150\).
• Combine like terms to simplify: \(6y + 35y - 68 = -150\) combines to \(41y - 68 = -150\).
• Solve for \(y\) and then back substitute to find \(x\).

Each step relies on basic algebraic rules such as the distributive property, combining like terms, and using inverse operations. Mastering these techniques is vital for efficiently solving equations by substitution or any other method.