Problem 14 - A Graphical Approach to College Algebra

Question

Solve each system by substitution. $$\begin{array}{r}2 x+y=-11 \\\x+3 y=-8\end{array}$$

Step-by-Step Solution

Answer

The solution to the system is \(x = -5\) and \(y = -1\).

Step 1: Solve for x in the Second Equation

Take the second equation, \(x + 3y = -8\). Solve for \(x\) by isolating it on one side of the equation. You can do this by subtracting \(3y\) from both sides, giving: \(x = -8 - 3y\).

Step 2: Substitute for x in the First Equation

Now, substitute the expression for \(x\) from Step 1 into the first equation, \(2x + y = -11\). Replace \(x\) with \(-8 - 3y\), so the equation becomes \(2(-8 - 3y) + y = -11\).

Step 3: Simplify and Solve for y

Expand and simplify the equation from Step 2: \(2(-8) + 2(-3y) + y = -11\), which becomes \(-16 - 6y + y = -11\). Combine like terms: \(-16 - 5y = -11\). Next, add \(16\) to both sides: \(-5y = 5\). Finally, divide by \(-5\): \(y = -1\).

Step 4: Substitute the Value of y back into x Expression

Now that we know \(y = -1\), substitute it back into the expression for \(x\) from Step 1: \(x = -8 - 3(-1)\). Simplify this by calculating \(-3(-1)\) to get \(x = -8 + 3 = -5\).

Step 5: Verify the Solution

Verify the solution \(x = -5\) and \(y = -1\) by plugging these values into the original equations. For the first equation: \(2(-5) + (-1) = -10 - 1 = -11\). For the second equation: \(-5 + 3(-1) = -5 - 3 = -8\). Both equations check out, confirming the solution.

Key Concepts

Substitution Method

When faced with solving systems of equations, one powerful technique that comes in handy is the substitution method. It involves a simple, systematic approach that can make complex systems more manageable. The first step in the substitution method is to solve one of the equations for one of the variables. This involves expressing one variable in terms of the other(s). For example, from the equation \(x + 3y = -8\), we can isolate \(x\) to get \(x = -8 - 3y\). This expression is now ready to be substituted into the other equation.

• Always choose the easiest equation and variable to isolate.
• Rearrange the equation to express one variable in terms of the others.

After substituting the isolated variable from step one into the other equation, you simplify and solve for the second variable. This helps to eliminate one of the unknowns, reducing it to a single-variable equation that is much easier to solve.

Linear Equations

Linear equations are equations of the first order. This means that they graph as straight lines and have no exponents higher than one. Linear equations are the bedrock of algebra and can often be found in systems of equations. They follow the standard form of \(ax + by + c = 0\) where \(a\), \(b\), and \(c\) are constants.

• Each equation in the system represents a line on a graph.
• The solution is the point where these lines intersect.

In our exercise, both equations \(2x + y = -11\) and \(x + 3y = -8\) are linear. The solution to a system of linear equations is essentially seeking the intersection point of the lines represented by these equations. By substituting and then solving, we find this intersection in algebraic terms.

Solving Systems

Solving systems of equations involves finding values for the variables that satisfy all equations simultaneously. There are various methods, such as graphing, substitution, and elimination, but our focus here is on substitution.

• The substitution method simplifies one equation to express a single variable, replacing it in the other equation.
• Simplification leads to a straightforward calculation of one variable, reducing complexity.
• Consequently, this solution is substituted back to find the remaining unknowns.

Finally, the solution must be verified by substituting the values of the variables back into the original equations. This ensures that both equations hold true, confirming the solution's correctness, as demonstrated by verifying that \(x = -5\) and \(y = -1\) satisfied both original equations in our example. Remember, understanding each method solidifies foundational algebra skills, allowing you to tackle more complex problems with confidence.