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Chapter 3: Problem 79

Determine \(a\) and \(b\) for which \((-4,-3)\) is a solution of the system $$ \begin{aligned} &a x+b y=-26\\\ &b x-a y=7 \end{aligned} $$

Short Answer

Expert verified
The values are \(a = 5\) and \(b = 2\).

Step by step solution

01

Substitute \((-4, -3)\) into the First Equation

Substitute the values \(x = -4\) and \(y = -3\) into the first equation \(a x + b y = -26\).\[-4a - 3b = -26\]
02

Simplify the First Equation

Rearrange the equation from Step 1 to solve for \(a\) in terms of \(b\).\[-4a - 3b = -26\ \Rightarrow -4a = -26 + 3b\ \Rightarrow a = \frac{26 - 3b}{4}\]
03

Substitute \((-4, -3)\) into the Second Equation

Substitute the values \(x = -4\) and \(y = -3\) into the second equation \(b x - a y = 7\).\[b(-4) - a(-3) = 7\ \Rightarrow -4b + 3a = 7\]
04

Solve for \(b\) and \(a\)

Substitute the expression for \(a\) from Step 2 into the equation from Step 3 and solve for \(b\).\[-4b + 3 \frac{26 - 3b}{4} = 7\ \Rightarrow -4b + \frac{3(26 - 3b)}{4} = 7\ \Rightarrow -4b + \frac{78 - 9b}{4} = 7\ \Rightarrow -16b + 78 - 9b = 28\ \Rightarrow -25b + 78 = 28\ \Rightarrow -25b = -50\ \Rightarrow b = 2\]\Substitute this \(b\) back into the expression for \(a\):\[a = \frac{26 - 3(2)}{4} = \frac{26 - 6}{4} = \frac{20}{4} = 5\]

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Key Concepts

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

Linear Systems
Understanding linear systems is fundamental to algebra. A linear system consists of two or more linear equations. These equations involve variables with no exponents or powers. The goal is to find the values of these variables that make all the equations true simultaneously.
In this exercise, we have a system of equations with two variables, \(a\) and \(b\), and we need to determine their values. The coordinates \((-4, -3)\) are given as a solution to this system, which tells us these values must satisfy both equations.
Substitution Method
The substitution method is a great way to solve systems of equations. First, you solve one of the equations for one variable. Then, you substitute this expression into the other equation.
Let's break down its application in our solution.
In our problem, we substitute \((-4, -3)\) into the first equation, giving us an equation in terms of \(a\) and \(b\). Next, we solve this new equation for \(a\) in terms of \(b\), which we鈥檒l use in the next steps.
Start by substituting \(x = -4\) and \(y = -3\) into the first equation:
\[-4a - 3b = -26\] We then solve for \(a\) in terms of \(b\):
\[a = \frac{26 - 3b}{4}\]
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to isolate variables. This is essential for solving systems of equations.
For the second part of our problem, we substitute \((-4, -3)\) into the second equation. Here鈥檚 how we do it:
Substituting in the second equation gives: \(-4b + 3a = 7\).
Using our earlier expression for \(a\), we have:
\[-4b + 3 \left(\frac{26 - 3b}{4}\right) = 7\]
Multiply everything out to solve for \(b\):\[-16b + 78 - 9b = 28\]
This simplifies to \(-25b + 78 = 28\).
Finally, we solve for \(b\): \[b = 2\].
Solution Verification
Verification is checking if your solution makes both equations true. This confirms you鈥檝e solved the system correctly.
We found \(b = 2\). Now substitute this back in our expression for \(a\): \(a = \frac{26 - 6}{4} = 5\).
To verify, substitute \(a = 5\) and \(b = 2\) back into the original system of equations:
First equation: \[5(-4) + 2(-3) = -20 - 6 = -26\]
Second equation: \[2(-4) - 5(-3) = -8 + 15 = 7\]
Both originals hold, meaning our solution \((a, b) = (5, 2)\) is correct.

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