Problem 36 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}4 a+2 b=-4 \\ 6 a-5 b=18\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \(\left(\frac{1}{2}, -3\right)\).

Step 1: Solve for one variable

To use the substitution method, we need to express one variable in terms of the other using one of the equations. Let's solve the first equation for \(b\).\[ 4a + 2b = -4 \] Subtract \(4a\) from both sides:\[ 2b = -4 - 4a \] Divide both sides by 2:\[ b = -2 - 2a \]

Step 2: Substitute into the second equation

Now that we have \(b\) in terms of \(a\), we substitute this expression into the second equation.\[ 6a - 5b = 18 \] Substitute \(b = -2 - 2a\):\[ 6a - 5(-2 - 2a) = 18 \]

Step 3: Simplify and solve for a

Distribute the \(-5\) in the equation:\[ 6a + 10 + 10a = 18 \] Combine like terms:\[ 16a + 10 = 18 \] Subtract 10 from both sides:\[ 16a = 8 \] Divide both sides by 16:\[ a = \frac{1}{2} \]

Step 4: Solve for b using the value of a

Now that we know \(a = \frac{1}{2}\), substitute this value back into the expression for \(b\).\[ b = -2 - 2a \] Plug in \(a = \frac{1}{2}\):\[ b = -2 - 2\left(\frac{1}{2}\right) \] Calculate:\[ b = -2 - 1 = -3 \]

Step 5: Write the solution as an ordered pair

Given that \(a = \frac{1}{2}\) and \(b = -3\), the solution to the system of equations is:\[ \left(\frac{1}{2}, -3\right) \]

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations that we solve together because they share variables. Each equation in a system describes a relationship between the variables. For example, think of it as a puzzle where each piece (equation) needs to fit in the right place to find the complete solution set for the variables. In the scenario where you're given multiple equations, you need to find the values of the variables that satisfy all equations simultaneously.

In this exercise, we have a system with two equations:

• First Equation: \(4a + 2b = -4\)
• Second Equation: \(6a - 5b = 18\)

The goal is to find values of \(a\) and \(b\) that make both equations true at the same time. The solution, expressed as an ordered pair, represents this set of values.

Linear Equations

Linear equations form the backbone of systems of equations. They are equations of the first degree, which means each term is linear (i.e., the variable is raised only to the power of one). You will see these equations commonly written in the form of \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants, and \(x\) and \(y\) represent variables.

In our exercise:

• The equation \(4a + 2b = -4\) is a linear equation with variables \(a\) and \(b\).
• Similarly, \(6a - 5b = 18\) is another linear equation.

Understanding the linear nature is crucial because it helps you recognize methods such as substitution that can solve these effectively.

Algebraic Solutions

Algebraic solutions involve manipulating equations to isolate the variable and find its value. The substitution method is one common algebraic technique used here. We isolate one of the variables in one equation and substitute it into the other equation. This transforms a system of equations into a single equation, which allows for easier solving.

The steps taken to solve the problem were straightforward:

• Rewrite the first equation, solving for \(b\) in terms of \(a\): \(b = -2 - 2a\).
• Substitute this expression into the second equation, replacing \(b\): \(6a - 5(-2 - 2a) = 18\).
• Solve the resulting equation for \(a\): \(a = \frac{1}{2}\).
• Substitute back the value of \(a\) into the expression for \(b\) to find its value: \(b = -3\).

This systematic approach leads us swiftly to the solution, which is \(\left(\frac{1}{2}, -3\right)\). These values satisfy both linear equations, thus solving the system.