Problem 38 - Elementary Algebra Within Reach

Question

In Exercises 35-38, solve the system by the method of elimination. $$ \left\\{\begin{aligned} \frac{1}{2}(x-4)+9 &=y-10 \\ -5(x+3) &=8-2(y-3) \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution for the system of equations is (0, 7).

Step 1: Expand the Expressions

First, we simplify the given system of equations by expanding the expressions and arranging it in the form of ax + by = c. \n\[\begin{aligned} \frac{1}{2}(x-4)+9 &=y-10 \ \frac{1}{2}x - 2 +9 &= y-10 \ \frac{1}{2}x + 7 = y \\ -5(x+3) &=8-2(y-3) \ -5x -15 &=8-2y +6 \ -5x -15 &= 14 - 2y \end{aligned}\]

Step 2: Rearrange the Second Equation

Next, we rearrange the second equation so that it fits the standard form (ax + by = c). We then get two equations with x and y on one side, and constant on the other. \n\[\begin{aligned} \frac{1}{2}x - y &= -7 \ -5x + 2y &= 14 \end{aligned}\]

Step 3: Find the Solution

Now, multiply the first equation by 2 to match the coefficient of y in the second equation: \n\[\begin{aligned} x - 2y &= -14 \ -5x + 2y &= 14 \end{aligned}\]\nThen add the two equations together to find x: \(x = 0\). Substitute x = 0 into the first equation to solve for y: \n\[-2y = -14 + 0 \rightarrow y = 7.\]

Step 4: Summary of the Solution

The solution to the system of equations is x = 0 and y = 7. This means that the two lines represented by the system intersect at the point (0, 7).

Key Concepts

Method of Elimination

The method of elimination is a powerful tool for solving systems of linear equations. This technique involves creating a situation where we can "eliminate" one of the variables. By manipulating the equations, we add or subtract them to cancel out one of the variables.

This method is highly effective when you can easily match the coefficients of one variable in both equations.

• First, take the given system and rewrite it to have like terms aligned vertically.
• Multiply one or both of the equations by a number that allows the coefficients of one variable in both equations to be the same or opposites.
• Add or subtract the equations to eliminate the said variable, resulting in a single equation with one unknown.
• Solve the remaining equation for the unknown variable.

Finally, use the value obtained from one variable to back-substitute into one of the original equations to find the other variable. Keep it simple by choosing the equation that appears easiest to manipulate.

Algebraic Expressions

Before adopting any method for solving equations, it's crucial to have a clear and straightforward form of the given algebraic expressions. Algebraic expressions consist of numbers, variables, and operations. They can be expanded, factored, or simplified depending on the context.

In our problem, we start with:

• An expression for one equation: \( \frac{1}{2}(x-4) + 9 = y - 10 \)
• Another expression: \(-5(x+3) = 8 - 2(y-3) \)

These initial forms are simplified by expanding to get statements with individual terms fully expressed.

Remember: Expansion multiplies out brackets, while simplification often involves aligning terms like variables on one side of the equation to help with further operations, such as elimination. Expanding expressions often makes it easier to identify what's needed to progress with solving.

Linear Equations

Linear equations are straightforward algebraic equations where each term is either a constant or the product of a constant with a single variable. A standard form of a linear equation in two variables looks like this: \( ax + by = c \).

These are called "linear" because their graphical representations are straight lines when plotted on a coordinate plane.

In the given exercise, both equations are carefully rewritten to fit this form:

• First equation: \( \frac{1}{2}x - y = -7 \)
• Second equation: \( -5x + 2y = 14 \)

The strategy with linear equations often involves obtaining equal coefficients for one variable to allow for elimination, making solving quite efficient. Remember, each solution set represents a point of intersection where the lines meet in the graph.

Solving Equations

Solving equations, especially systems of them, involves finding the set of values that satisfy all given conditions. After aligning the system in the standard linear form, solving involves strategic manipulation.

In our exercise, solving happens in layers:

• First, manipulate to aid elimination by adjusting a variable's coefficients.
• Add or subtract equations to effectively remove one variable, leaving an equation with one unknown.
• Substitute the result back to find the second variable.

The solving step can also include verifying the solutions by plugging the values back into the original equations. For example, substituting \( x = 0 \) into the chosen equation yields \( y = 7 \). It's also feasible to check both by verifying with the entire system.

This verification step ensures the solution is accurate, especially crucial when working through homework or exams, where confidence in answers is key.