Problem 38 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} 16 x-8 y=5 \\ 32 x+8 y=19 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = \frac{9.5}{16}\) and \(y = \frac{9}{16}\)

Step 1: Express one variable in terms of the other in the first equation

From the first equation \(16x - 8y = 5\), let's express \(x\) in terms of \(y\). We do this by first doing \(16x = 5 + 8y\) and then dividing by \(16\) to isolate \(x\). We get \(x = \frac{5 + 8y}{16}\).

Step 2: Substitute expression from Step 1 into the second equation

We substitute the expression \(x = \frac{5 + 8y}{16}\) from Step 1 into the second equation. That gives us \(32(\frac{5 + 8y}{16}) + 8y = 19\). After cleaning this up, we get \(8y + 8y = 19 - 10\), and upon further simplifying, we have \(16y = 9\). Finally, solving for \(y\) gives us \(y = \frac{9}{16}\).

Step 3: Substitute \(y\) value into the first equation to find \(x\)

Now we substitute \(y = \frac{9}{16}\) into the first equation \(16x - 8y = 5\). That gives us \(16x - 8(\frac{9}{16}) = 5\). Simplifying this leads to \(16x - 4.5 = 5\), further simplification gives us \(16x = 9.5\), and isolating \(x\) gives \(x = \frac{9.5}{16}\).

Key Concepts

system of equations

A system of equations consists of two or more equations that share common variables. In solving these, the goal is to find a set of values for the variables that make all the equations true simultaneously. The given system of equations in this exercise can be represented as:

• Equation 1: \(16x - 8y = 5\)
• Equation 2: \(32x + 8y = 19\)

Each of these equations represents a line on a graph. Finding the solution to the system means finding the point where these lines intersect, if they do. Using methods like substitution or elimination helps determine that intersection point by focusing on relationships between the variables.

solving algebra equations

Solving algebra equations involves manipulating the equation to find the values of the unknown variables. The goal is to isolate the unknown variable on one side of the equation. Techniques such as addition, subtraction, multiplication, division, and substitution are used.In our problem, we used substitution to simplify one equation and remove one variable, making it easier to solve for the other. For instance, expressing \(x\) in terms of \(y\) converts our system of equations into a form that allows straightforward computations. This linear manipulation is a standard approach to tackling problems involving systems of equations.

linear equations

Linear equations are equations of the first degree, which means they involve variables raised only to the power of one. They are represented in the standard form \(ax + by = c\) and graphically appear as straight lines.In the system given, each equation aligns with this structure:

• For \(16x - 8y = 5\), the coefficients 16 and -8 dictate the line's slope and position.
• For \(32x + 8y = 19\), we see a different orientation as they create another line intersecting or potentially going parallel or perpendicular to the first.

These equations play a crucial role in determining their graphical solutions and in many real-world scenarios, such as predicting and modeling outcomes.

variable isolation

Variable isolation is a technique used to rearrange an equation so that the variable of interest is alone on one side of the equation. This is central to solving for unknowns in algebraic equations.In our exercise, during Step 1, variable isolation involved expressing \(x\) in terms of \(y\) from the first equation: - Start with: \(16x - 8y = 5\)- Rearrange to isolate \(x\): First add \(8y\) to both sides, then divide by 16- Resulting in: \(x = \frac{5 + 8y}{16}\)This was a deliberate attempt to simplify and subsequently resolve the value of \(y\), enabling straightforward computation of \(x\) in future steps. Such techniques simplify complex systems by reducing them step-by-step.