Problem 66 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{4 x+y=7} \\ {3 x+\frac{4}{5} y=5.5}\end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 0.5\) and \(y = 5\).

Step 1: Rearrange First Equation in Terms of y

Start with the first equation: \(4x + y = 7\). Solve for \( y\) by subtracting \(4x\) from both sides: \(y = 7 - 4x\).

Step 2: Substitute y in Second Equation

Take the expression for \( y\) from the first equation and substitute it into the second equation: \(3x + \frac{4}{5}(7 - 4x) = 5.5\).

Step 3: Distribute and Simplify

Distribute \(\frac{4}{5}\) in the second equation: \(3x + \frac{28}{5} - \frac{16}{5}x = 5.5\). Simplify to \((3 - \frac{16}{5})x + \frac{28}{5} = 5.5\) which becomes \(\frac{15}{5} - \frac{16}{5} = \frac{-1}{5}\). Therefore, the equation is \(\frac{-1}{5}x + \frac{28}{5} = 5.5\).

Step 4: Solve for x First

Rearrange the equation \(\frac{-1}{5}x + \frac{28}{5} = 5.5\) to find \(x\). Multiply all terms by 5 to eliminate the fraction: \(-x + 28 = 27.5\). Solve for \(x\) by subtracting 28 from both sides: \(-x = -0.5\). Thus, \(x = 0.5\).

Step 5: Solve for y using x Value

Substitute \(x = 0.5\) back into the expression for \(y\) from Step 1: \(y = 7 - 4(0.5)\). Simplify to get \(y = 7 - 2 = 5\).

Key Concepts

Substitution Method

The substitution method is a useful technique for solving systems of equations, especially when one of the equations is easy to manipulate. This method involves solving one equation for one variable and then substituting that expression into the other equation(s). By doing this, the system of equations is reduced to a single equation with one unknown, which can arguably simplify the problem.

In the original exercise, the first equation was given as \(4x + y = 7\). This equation was rearranged to express \(y\) in terms of \(x\): \(y = 7 - 4x\). This step is crucial because it allows us to substitute \(7 - 4x\) in place of \(y\) in the second equation.

By substituting into the second equation, \(3 x + \frac{4}{5} y = 5.5\), the system is simplified into one equation \(3x + \frac{4}{5}(7 - 4x) = 5.5\). Through substitution, the problem becomes more straightforward as we now focus on solving a single equation for the variable \(x\), leading us closer to the solution.

Linear Equations

Linear equations are fundamental in algebra and are characterized by having variables raised only to the first power. In a system of linear equations like the one given, each equation represents a straight line when graphed on a coordinate plane.

For the provided system, we have two linear equations:

• \(4x + y = 7\)
• \(3x + \frac{4}{5}y = 5.5\)

Each equation can be viewed as a line, and solving the system means finding the point where these two lines intersect.

Since the solutions to the system are the values of \(x\) and \(y\) that satisfy both equations simultaneously, their intersection yields the pair \((x, y)\). This point of intersection is often the solution you seek when handling systems of linear equations. For our specific system, the lines intersect at \((0.5, 5)\).

Solving Systems Algebraically

Solving systems algebraically, as opposed to graphically or numerically, involves calculating precise answers through methods such as substitution or elimination. The goal is to find a set of variable values that satisfy all equations in the system.

The substitution method, as demonstrated in the solution, is a potent algebraic technique. After substituting \(y = 7 - 4x\) into the second equation, we simplified to a single equation with one unknown: \(\frac{-1}{5}x + \frac{28}{5} = 5.5\).

This equation was then solved by eliminating fractions—multiplying through by 5, resulting in \(-x + 28 = 27.5\). Solving for \(x\) yields \(x = 0.5\).
Subsequently, this value was substituted back to find \(y\), confirming \(y = 5\). This method beautifully illustrates how algebra helps in systematically solving equations step by step, leading to an exact solution \((0.5, 5)\) for the system. Through algebra, especially when dealing with linear equations and systems, precision in solving similar problems is effectively achieved.