Problem 12 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} x+3 y=-5 \\ 2 x+2 y=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 7\), \(y = -4\).

Step 1: Solve the first equation for x

Let's begin by isolating \(x\) in the first equation \(x + 3y = -5\). To do this, subtract \(3y\) from both sides of the equation: \[ x = -5 - 3y \]

Step 2: Substitute expression for x into the second equation

Use the expression for \(x\) from Step 1 and substitute it into the second equation \(2x + 2y = 6\): \[ 2(-5 - 3y) + 2y = 6 \] which simplifies to: \[ -10 - 6y + 2y = 6 \]

Step 3: Simplify and solve for y

Now, simplify the equation: \[ -10 - 4y = 6 \]Add \(10\) to both sides to isolate terms with \(y\): \[ -4y = 16 \]Next, divide by \(-4\) to solve for \(y\): \[ y = -4 \]

Step 4: Substitute y back to find x

Use the value of \(y\) to find \(x\). Substitute \(y = -4\) back into the equation \(x = -5 - 3y\): \[ x = -5 - 3(-4) \] Simplify this to: \[ x = -5 + 12 \] Which results in: \[ x = 7 \]

Step 5: Verify the solution

Substitute \(x = 7\) and \(y = -4\) back into the original equations to ensure they are satisfied:For the equation \(x + 3y = -5\), substitute:\[ 7 + 3(-4) = 7 - 12 = -5 \] which holds true.For the equation \(2x + 2y = 6\), substitute:\[ 2(7) + 2(-4) = 14 - 8 = 6 \] which also holds true. Thus, the solution \((x, y) = (7, -4)\) is correct.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In our example, the system consists of two equations with variables \(x\) and \(y\): \[ \begin{aligned} x + 3y &= -5 \ 2x + 2y &= 6 \end{aligned} \]The primary goal when working with a system of equations is to find a solution that satisfies all equations simultaneously. This means finding values for the variables that make all equations true at the same time.
In solving systems like these, various methods can be used, such as substitution, elimination, and graphing. Each of these methods has its own advantages and fits different situations better, depending on the specific system you are working with.

Linear Equations

Linear equations are mathematical expressions that describe a straight line when graphed on a coordinate plane. These equations have no exponents higher than one, making them straightforward to work with. The structure of a linear equation in two variables typically looks like this:\[ ax + by = c \]where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables.
In our exercise, both equations given are linear: - \( x + 3y = -5 \)- \(2x + 2y = 6 \)Both have variables raised only to the power of one. These equations can be plotted as lines on a graph, and their intersection would be the point that is the solution to the system of equations.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. In the context of a system of equations, "solving" means identifying a pair of values that satisfy both equations simultaneously.
The substitution method, used in the exercise, involves:

• Isolating one variable: The first step is to solve for one of the variables in one of the equations. This is done by expressing one variable in terms of another, as shown when we isolated \(x\) with \(x = -5 - 3y\).
• Substituting the value: Next, substitute this expression into the other equation, transforming it into a single-variable equation that is easier to solve.
• Solving the equation: Simplify and solve for the isolated variable. Here, we solved for \(y\), yielding \(y = -4\).
• Back-substituting: Finally, substitute back to find the other variable, in this instance giving us \(x = 7\).

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to make them easier to solve. This can involve operations such as adding, subtracting, multiplying, or dividing terms, as well as factoring or expanding expressions.
In the provided solution, algebraic manipulation was crucial in several steps:

• Subtracting \(3y\) from both sides to isolate \(x\) in the first equation.
• Expanding and simplifying the substituted equation \(2(-5 - 3y) + 2y = 6\) to \(-10 - 4y = 6\).
• Combining like terms and solving for \(y\).
• Substituting back to solve for \(x\) and checking that the found values satisfy both equations.

These manipulations are foundational skills in algebra, enabling us to systematically approach and solve even more complex systems of equations.