Problem 98 - Elementary Algebra

Question

$$ \left\\{\begin{array}{l} 2 x-15 y=7 \\ 12 x+2 y=-4 \end{array}\right. $$$$ \left\\{\begin{array}{l} 2 x-15 y=7 \\ 12 x+2 y=-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{1}{4}, y = -\frac{1}{2}\).

Step 1: Express one variable in terms of the other

From the first equation, solve for one of the variables. Let's solve for x.\(2x - 15y = 7\). Hence, \(2x = 7 + 15y\) which gives \(x = \frac{7 + 15y}{2}\).

Step 2: Substitute the expression into the second equation

Now, substitute the expression for x into the second equation.\(12x + 2y = -4\), substituting \(x = \frac{7 + 15y}{2}\), we get\[12\bigg(\frac{7 + 15y}{2}\bigg) + 2y = -4\] Simplify to get\[6(7 + 15y) + 2y = -4\].

Step 3: Simplify and solve for y

Simplify the equation to solve for y:\[42 + 90y + 2y = -4\]. Combine like terms:\[42 + 92y = -4\]. Then solve for y:\[92y = -46\], yielding \(y = -\frac{46}{92} = -\frac{1}{2}\).

Step 4: Substitute y back to find x

Now, substitute \(y = -\frac{1}{2}\) back into the expression for x:\[x = \frac{7 + 15y}{2}\ = \frac{7 + 15\big(-\frac{1}{2}\big)}{2}\ = \frac{7 - \frac{15}{2}}{2}\ = \frac{14 - 15}{4}\ = -\frac{1}{4}\].

Step 5: Verify the solution

Substitute \(x = -\frac{1}{4} \) and \(y = -\frac{1}{2} \) into the original equations to check the solution is accurate.

Key Concepts

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. They are called 'linear' because they graph as straight lines on a coordinate plane. In general, a linear equation in two variables looks like this:
\[ ax + by = c \]
In this exercise, we have two linear equations:

• \( 2x - 15y = 7 \)
• \( 12x + 2y = -4 \)

These equations form a system, which can be solved using various methods. One such method is the substitution method, which we will explore further.

Substitution Method

The substitution method is a technique for solving systems of equations. The main idea is to solve one equation for one of the variables and then substitute this expression into the other equation. This reduces the system to a single equation in one variable, which is easier to solve.
Let's break this down with our given system. First, we solve the first equation for \(x\):
\[ 2x - 15y = 7 \]
Isolate \(x\) by adding \(15y\) to both sides and then dividing by 2:
\[ 2x = 7 + 15y \]\[ x = \frac{7 + 15y}{2} \]
Now we take this expression for \(x\) and substitute it into the second equation, resulting in a single equation with one variable. The substitution simplifies the system, and makes it easier to solve for the remaining variables.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and expressions to isolate variables or simplify the problem. This is a crucial step in the substitution method.
After substituting \( x = \frac{7 + 15y}{2} \) into the second equation, we get:
\[ 12 \bigg( \frac{7 + 15y}{2} \bigg) + 2y = -4 \]
Simplify inside the parentheses first:
\[ 6(7 + 15y) + 2y = -4 \]
Distribute the 6:
\[ 42 + 90y + 2y = -4 \]
Combine like terms:
\[ 42 + 92y = -4 \]
Next, isolate \(y\) by subtracting 42 from both sides:
\[ 92y = -46 \]
Finally, divide by 92:
\[ y = -\frac{46}{92} = -\frac{1}{2} \]
These steps of algebraic manipulation help us simplify the problem and find \(y\).

Solving for Variables

After isolating and solving for \(y\), we next solve for \(x\). To do this, we substitute the value of \(y\) back into our earlier expression for \(x\):
\[ x = \frac{7 + 15(-\frac{1}{2})}{2} = \frac{7 - \frac{15}{2}}{2} \] \[ x = \frac{14 - 15}{4} = -\frac{1}{4} \]
Now, we have our solution: \( x = -\frac{1}{4} \) and \( y = -\frac{1}{2} \).
It is always good practice to verify the solution by substituting these values back into the original equations. If they satisfy both equations, then the solution is correct:

• For \( 2x - 15y = 7 \): Verify by substituting \( x \) and \( y \).
• For \( 12x + 2y = -4 \): Verify by substituting \( x \) and \( y \).

Both should hold true, confirming that our solution is accurate.