Problem 37 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}5 a+6 b=8 \\ 2 a-15 b=9\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \(a = 2\) and \(b = -\frac{29}{87}\).

Step 1: Solve for one variable in one equation

Let's isolate variable \(a\) from the first equation: 5a + 6b = 8. Solve for \(a\): \[ a = \frac{8 - 6b}{5} \].

Step 2: Substitute into the second equation

Take the expression for \(a\) from Step 1 and substitute it into the second equation: 2a - 15b = 9. Replace \(a\) with \(\frac{8 - 6b}{5}\): \[ 2\left(\frac{8 - 6b}{5}\right) - 15b = 9 \].

Step 3: Simplify the substituted equation

Simplify the equation from Step 2: \[ \frac{16 - 12b}{5} - 15b = 9 \].Multiply every term by 5 to eliminate the fraction: \[ 16 - 12b - 75b = 45 \].

Step 4: Solve for the remaining variable

Combine like terms: \[-87b = 45 - 16 \],which simplifies to: \[-87b = 29 \].Divide by -87 to solve for \(b\): \( b = \frac{29}{-87} \),or \( b = -\frac{29}{87} \).

Step 5: Substitute back to find the other variable

Now use the expression for \(b\) to find \(a\). Substitute \(b = -\frac{29}{87}\) back into the expression for \(a\) from Step 1:\[ a = \frac{8 - 6(-\frac{29}{87})}{5} \].Simplify:\[ a = \frac{8 + \frac{174}{87}}{5} \].Convert \(8\) to an equivalent fraction:\[ a = \frac{\frac{696}{87} + \frac{174}{87}}{5} = \frac{870}{87} \].Hence, \[ a = \frac{870}{435} = 2 \].

Step 6: Verify the solution

Substitute \( a = 2 \) and \( b = -\frac{29}{87} \) back into the original equations to verify:For the first equation, \[5(2) + 6\left(-\frac{29}{87}\right) = 8\]Calculate:\[10 - \frac{174}{87} = 8 \]which simplifies to \[10 - 2 = 8 \].The solution fits the equation correctly. For the second equation:\[2(2) - 15\left(-\frac{29}{87}\right) = 9\]Calculate:\[4 + \frac{435}{87} = 9 \]which simplifies to \[4 + 5 = 9 \].This also fits. The solution is verified.

Key Concepts

Solving Systems of Equations Using the Substitution Method

The substitution method is a systematic approach to solve systems of linear equations, like the one given. The goal is to find values for variables that satisfy all the equations in the system. Here, we have two equations involving variables \( a \) and \( b \):

• First equation: \( 5a + 6b = 8 \)
• Second equation: \( 2a - 15b = 9 \)

The substitution method typically involves the following steps:

• Isolate a Variable: First, select one equation and solve for one of the variables. In our case, we isolated \( a \) in the first equation: \( a = \frac{8 - 6b}{5} \).
• Substitute: Next, take the expression found for \( a \) and substitute it into the other equation. This step helps in reducing two-variable equations to a single-variable equation.

By substituting the expression for \( a \) into the second equation, we create a situation where we only need to solve for \( b \). This reduces complexity and makes the process more manageable.

Algebraic Manipulation to Simplify and Solve Equations

Algebraic manipulation involves rearranging equations to isolate and solve for unknown variables. This requires skillful handling of algebraic expressions, following arithmetic operations, and simplification techniques.In the context of the substitution method, algebraic manipulation is crucial at several stages:

• Simplifying the Substituted Equation: After substituting the expression for \( a \), the resulting equation \( \frac{16 - 12b}{5} - 15b = 9 \) needs simplification. Multiply through by 5 to remove fractions: \( 16 - 12b - 75b = 45 \).
• Combining Like Terms: Gather like terms to simplify: \(-87b = 45 - 16\), resulting in \(-87b = 29\). This step reduces the equation to a simple format \(-87b = 29\), ready for solving.
• Solving for a Variable: Divide or multiply throughout by the appropriate term to solve for the variable: \( b = -\frac{29}{87} \).

Algebraic manipulation requires careful attention to detail. Each step ensures equations are handled correctly, leading us to a valid solution for the system.

Verifying Solutions through Substitution

Verifying solutions is an essential step to ensure accuracy and correctness in solving algebraic equations. It involves plugging the found values of variables back into the original equations to check if they hold true.Once solutions for \( a \) and \( b \) are determined, verify them like so:

• First Equation: Substitute \( a = 2 \) and \( b = -\frac{29}{87} \) into \( 5a + 6b = 8 \). Calculate to check: \( 10 - \frac{174}{87} = 8 \), thus simplifying to \( 8 = 8 \).
• Second Equation: Use the same values in \( 2a - 15b = 9 \). Calculate \( 4 + \frac{435}{87} \), which simplifies to \( 9 = 9 \).

Both checks confirm that our solutions are correct. Always verify your solutions, as it guarantees that they fulfill all conditions of the original system. This practice ensures reliability and accuracy in problem-solving.