Problem 37 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} 3 x+2 y=12 \\ x-y=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 6\) and \(y =\frac{3}{5}\).

Step 1: Isolate a Variable in One Equation

Firstly, one needs to isolate a variable in one of the equations. Let’s isolate \(x\) in the second equation: \[x = y + 3\]

Step 2: Substitute into Other Equation

Now we can substitute the expression obtained for \(x\) into the first equation: \[3(y + 3) + 2y = 12\]

Step 3: Solve for Variable

Next, solve the equation obtained for \(y\). First, simplify the equation: \[3y + 9 + 2y = 12 \Rightarrow 5y + 9 = 12\]. Then, subtract 9 from both sides to solve for \(y\): \[5y = 12 - 9 \Rightarrow 5y = 3 \Rightarrow y = \frac{3}{5}\]

Step 4: Substitute Back to Get the Other Variable

Now substitute the computed \(y\)-value into the isolated equation from step 1, to solve for \(x\): \[x = \frac{3}{5} + 3 = \frac{15}{5} + \frac{3}{1} = 3 + 3 = 6\]

Key Concepts

Substitution Method

The substitution method is like solving a puzzle piece by piece. When dealing with systems of equations, this approach lets us find values for unknown variables. Here’s how it works: you solve one of the equations for one variable, and then substitute that expression into the other equation. This simplification reduces the problem to a single equation with one variable, which can be dealt with more easily. This technique is particularly useful when the system includes a simple equation or when one of the coefficients is one, making isolation and substitution less complex. It’s like peeling off the outer layers to get to the core of the problem and solve it effectively.

Isolating Variables

Isolating variables is a critical step in solving equations through substitution. This means rearranging an equation so that one variable is by itself on one side of the equation. For example, if we have an equation like

• \(x - y = 3\),

we can isolate \(x\) by adding \(y\) to both sides, leading to

• \(x = y + 3\).

This step is essential because it sets up the next stage, where the expression for \(x\) (in terms of \(y\)) is substituted into another equation. Isolating a variable simplifies the process, cutting down on the complexity by dealing with one less variable in the main equation for the time being.

Solving Linear Equations

After substitution, the next focus shifts to solving linear equations. A linear equation will look something like \(ax + by = c\), with \(a\), \(b\), and \(c\) being constants. The beauty of linear equations lies in their simplicity and predictability. Once substituted, if you get an equation like

• \(5y + 9 = 12\),

solve by rearranging the equation to find \(y\). You start by getting rid of constants, for instance by subtracting 9 from both sides leading to

• \(5y = 3\),

and then dividing both sides by 5, resulting in

• \(y = \frac{3}{5}\).

This systematic approach ensures you arrive at the solution accurately and efficiently.

Algebraic Substitution

Algebraic substitution is all about cleverly maneuvering variables between equations. Once you have isolated a variable from one equation, like expressing \(x\) in terms of \(y\), insert that expression into the other equation. This shifts the problem of multiple equations, each with multiple unknowns, into a more familiar one-variable equation. Consider inserting

• \(x = y + 3\)

into the other equation. This changes

• \(3x + 2y = 12\)

to

• \(3(y + 3) + 2y = 12\).

Simplifying further reduces it to a single-variable equation that you can solve. This step is like transforming a complex forest of equations into a clear path, guiding you straight to the answer.