Problem 50 - College Algebra

Question

Solve each system for \(x\) and \(y,\) expressing either value in terms of a or \(b,\) if necessary. Assume that \(a \neq 0\) and \(b \neq 0\). \(\left\\{\begin{array}{l}4 a x+b y-3 \\ 6 a x+5 b y-8\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions for \(x\) and \(y\) obtained will be in terms of \(a\) and \(b\). The exact answers depend on the detailed solutions in the steps outlined above.

Step 1: Express \(y\) as function of \(x\)

From the first equation express \(y\) in terms of \(x\), \(a\), and \(b\). This is done by rearranging the equation to \(y\) on one side: \(y = \frac{3 - 4ax}{b}\)

Step 2: Substitution in second equation

Substitute equation in Step 1 into the second equation in the system: \(6ax + 5b\left(\frac{3 - 4ax}{b}\right) = 8\). Simplify this to get an equation solely in terms of \(x\), \(a\), and \(b\).

Step 3: Solve for \(x\)

Solve the simplified equation from Step 2 for \(x\) to express \(x\) in terms of \(a\) and \(b\).

Step 4: Substitute \(x\) back into first equation

Substitute the solved \(x\) from Step 3 back into the first original equation in the system to get an equation solely in terms of \(y\), \(a\), and \(b\).

Step 5: Solve for \(y\)

Solve the equation from Step 4 for \(y\) to express \(y\) in terms of \(a\) and \(b\).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a key skill when dealing with system of equations. It involves rearranging and transforming expressions to isolate specific variables. This process often requires addition, subtraction, multiplication, or division to both sides of an equation to preserve equality. In the given system of equations, you manipulate the first equation to express the variable \(y\) in terms of \(x\), \(a\), and \(b\). Rearranging the equation to get \(y\) alone means you need to isolate \(y\) on one side of the equation. To do this, start by moving terms not involving \(y\) to the other side. Then, divide by the coefficient of \(y\). This results in \(y = \frac{3 - 4ax}{b}\). Use this to facilitate further steps such as substitution.

Substitution Method

The substitution method is an effective strategy for solving systems of equations. It involves taking an expression from one equation for a particular variable and substituting it into another equation. This reduces the system to a single equation with one variable. For this problem, you substitute \(y = \frac{3 - 4ax}{b}\) into the second equation. This step eliminates \(y\) from the second equation, allowing you to focus on solving for \(x\). After substitution, the equation becomes \(6ax + 5b\left(\frac{3 - 4ax}{b}\right) = 8\).Simplifying this expression involves removing parentheses and combining like terms, yielding an equation with only \(x\), \(a\), and \(b\). This streamlined equation can then be further manipulated to find the value of \(x\).

Solving Variables

Solving for variables is a crucial step in working out systems of equations. Once you reduce your equations to a single variable, it's time to "solve" or figure out the numerical or parameter-dependent value of that variable. In our solution process, starting with the simplified equation from substitution, you solve for \(x\). The goal is to rearrange the expression such that \(x\) is isolated. This might involve combining like terms, factoring when possible, or performing inverse operations like division or subtraction. Once \(x\) is found in terms of \(a\) and \(b\), you can take this solution and backtrack to determine the value of \(y\) by substituting \(x\) back into the expression \(y = \frac{3 - 4ax}{b}\) obtained from step 1.

Expressing in Terms of Parameters

Expressing variables in terms of parameters involves identifying relationships between variables and parameters, such as \(a\) and \(b\). This is especially useful when specific numerical values aren’t given for all elements in an equation.In the given exercise, since \(aeq 0\) and \(beq 0\), the solutions for \(x\) and \(y\) won't be specific numbers but rather expressions. These will include the parameters \(a\) and \(b\). After successfully solving for \(x\) first, we express \(x\) as functions involving \(a\) and \(b\). Following suit, you use the solved \(x\) terms back in the re-arranged initial equation to express \(y\) also in terms of \(a\) and \(b\). This gives a more flexible solution applicable to many scenarios. Being able to express solutions in terms of parameters is a powerful tool in algebra as it allows one to understand how change in one parameter affects the values of the variables.