Problem 23 - Thinking Mathematically

Question

Solve each system by the substitution method. Be sure to check all proposed solutions. \(\left\\{\begin{array}{rr}x+8 y= & 6 \\ 2 x+4 y= & -3\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -4\) and \(y = 1.25\).

Step 1: Isolate a variable from one equation

Choose one equation and isolate it for a variable. Here, the first equation \(x + 8y = 6\) can be rewritten as \(x = 6 - 8y\).

Step 2: Substitute into the second equation

Substitute \(x = 6 - 8y\) into the second equation to get \(2(6 - 8y) + 4y = -3\). Simplify this equation to find the value of \(y\). After simplifying, the equation is \(12 - 16y + 4y = -3\) which simplifies further to \(12 -12y = -3 \) and then to \(-12y = -3 - 12\) which gives \(-12y = -15\). Finally, we can find \(y\) by dividing both sides of the equation by \(-12\), which gives \(y = 15/12 = 1.25\).

Step 3: Substitute \(y\) back into the first equation

Substitute \(y = 1.25\) back into the first equation to find \(x\). This give us: \(x = 6 - 8(1.25) = 6 -10 = -4\).

Step 4: Check the solution

Substitute \(x = -4\) and \(y = 1.25\) into the original equations to verify the solution. The first equation gives \(-4 + 8(1.25) = -4 +10 = 6\). So the first equation is satisfied. For the second equation: \(2(-4) + 4(1.25) = -8 + 5 = -3\), which means the second equation is also satisfied. This verifies that the solutions \(x = -4\) and \(y = 1.25\) are correct.

Key Concepts

Solving Systems of Equations

When dealing with systems of equations, our goal is to find values for the variables that satisfy all the given equations simultaneously. Think of it like finding the intersection point of two lines on a graph. Sometimes lines meet at a single point, and that's where we find our solution.

There are several methods to solve systems of equations, but today we'll focus on the substitution method. The substitution method is especially useful when one of the equations is easy to solve for a single variable:

• Begin by isolating one variable in one of the equations. Choose the equation where isolation seems most straightforward.
• Substitute this expression into the other equation, allowing you to solve for the second variable.
• Once you have one value, plug it back into the expression from the first step to find the other variable.

By methodically following these steps, you can effectively solve systems of linear equations. Let's move on to understanding linear equations to give you clearer picture.

Linear Equations

Linear equations are foundational in mathematics and appear as straight lines on a graph. Each equation in its simplest form will look like ax + by = c, where x and y are variables, and a, b, and c are constants. The coefficients a and b determine the slope and direction of the line.

In a two-variable system, each equation graphically represents a line in a plane. The intersection of these lines—if any—indicates the solution to the system of equations. When using substitution, we make one equation express x in terms of y, or vice versa, allowing us to replace variables effectively.

It's crucial to understand that the simplicity of linearity lets us approach systems of equations with confidence. Once you grasp the properties of linear equations, solving them with methods like substitution becomes a lot easier and fun!

Mathematical Verification

After finding what might be the solution, it is essential to verify whether these values genuinely satisfy all equations in the system. Mathematical verification is the process of plugging back the found values into the original equations:

• Insert the solution of x and y back into the initial equations.
• Ensure that both sides of equations balance, confirming that the left-hand side equals the right-hand side.
• Repeat this for all the equations in your system.

This step is crucial as it ensures that no errors occurred along the way and guarantees that the solution is indeed correct. Always remember, a solution isn't valid until it has been verified to work in all the original equations without fail. This practice not only confirms your answer but enhances your understanding of algebraic manipulation.