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Chapter 11: Problem 17

Solve each system by using the substitution method. \(\left(\begin{array}{l}a=4 b+13 \\ 3 a+6 b=-33\end{array}\right)\)

Short Answer

Expert verified
The solution is \( a = -3 \) and \( b = -4 \).

Step by step solution

01

Solve the First Equation for 'a'

The first equation is given as \( a = 4b + 13 \). This is already solved for \( a \), so we can use this expression directly in the next step.
02

Substitute 'a' in the Second Equation

Substitute the expression we found for \( a \) into the second equation. Replace \( a \) in \( 3a + 6b = -33 \) with \( 4b + 13 \). This gives:\[3(4b + 13) + 6b = -33\]
03

Simplify the Equation

Expand and simplify the equation from Step 2:\[3(4b + 13) + 6b = 12b + 39 + 6b = -33\]Combine like terms:\[12b + 6b + 39 = -33\]\[18b + 39 = -33\]
04

Solve for 'b'

Subtract 39 from both sides of the equation:\[18b = -33 - 39\]\[18b = -72\]Divide by 18:\[b = -4\]
05

Solve for 'a' Using 'b'

Now that we have \( b = -4 \), substitute this back into the first equation \( a = 4b + 13 \):\[a = 4(-4) + 13\]\[a = -16 + 13\]\[a = -3\]
06

Verify the Solution

Substitute \( a = -3 \) and \( b = -4 \) back into both original equations to verify:- For the first equation: \( -3 = 4(-4) + 13 \) simplifies to \( -3 = -16 + 13 \), which holds true.- For the second equation: \( 3(-3) + 6(-4) = -33 \) simplifies to \( -9 - 24 = -33 \), which holds true. Thus, our solution is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The System of Equations
In algebra, a system of equations consists of two or more equations with the same set of variables. By solving a system of equations, we are finding the values of the variables that satisfy all the equations simultaneously. In the given exercise, our system of equations is:
  • \( a = 4b + 13 \)
  • \( 3a + 6b = -33 \)
This is known as a linear system because each equation represents a straight line when graphed. The intersection of the lines, or the point where they meet, is the solution to the system. In our case, we are interested in solving the system using the substitution method.
Understanding Linear Equations
Linear equations are the foundation of solving systems by the substitution method. Each equation in our system can be written in the general form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables.

In the exercise, the equations can be seen as relationships between \( a \) and \( b \). They describe planes in a two-dimensional space. When solving linear systems, our aim is to find values for \( a \) and \( b \) that satisfy both linear relationships at the same time.
But unlike graphical methods that require plotting, solving algebraically using substitution provides a precise solution without needing a graph.
  • Equation 1, \( a = 4b + 13 \), easily isolates \( a \) in terms of \( b \).
  • Equation 2, \( 3a + 6b = -33 \), uses these variables in combination.
Algebra Problem Solving with Substitution Method
The substitution method is a straightforward technique for solving systems of equations. Here's how it works step by step:
  • **Isolate**: First, pick one equation and isolate one variable. In our exercise, the first equation is already solved for \( a \) (\( a = 4b + 13 \)).
  • **Substitute**: Substitute this expression for the isolated variable into the other equation. Here, we substitute \( 4b + 13 \) for \( a \) in the second equation: \( 3(4b + 13) + 6b = -33 \)
  • **Solve**: Solve the resulting equation for the remaining variable. Simplifying gives \( b = -4 \).
  • **Back-substitute**: Finally, substitute back to find the other variable. Use \( b = -4 \) in \( a = 4b + 13 \) to find \( a = -3 \).
This technique is powerful because it systematically reduces the complexity of the problem, making seemingly complex solutions attainable by breaking them into simpler steps.

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Most popular questions from this chapter

Use Cramer's rule to find the solution set for each system. If the equations are dependent, simply indicate that there are infinitely many solutions. \(\left(\begin{array}{c}2 x-y+3 z=-5 \\ 3 x+4 y-2 z=-25 \\\ -x+z=6\end{array}\right)\)

Evaluate each \(3 \times 3\) determinant. Use the properties of determinants to your advantage. \(\left|\begin{array}{rrr}-3 & -2 & 1 \\ 5 & 0 & 6 \\ 2 & 1 & -4\end{array}\right|\)

Use Cramer's rule to find the solution set for each system. If the equations are dependent, simply indicate that there are infinitely many solutions. \(\left(\begin{array}{c}2 x+7 y=-1 \\ x=2\end{array}\right)\)

Consider the following matrix: $$ A=\left[\begin{array}{rrrr} 2 & 5 & 7 & 9 \\ -4 & 6 & 2 & 4 \\ 6 & 9 & 12 & 3 \\ 5 & 4 & -2 & 8 \end{array}\right] $$ Form matrix \(B\) by interchanging rows 1 and 3 of ma\(\operatorname{trix} A\). Now use your calculator to show that \(|B|=-|A|\).

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{-2 x-8}{x^{2}-1}\)
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