Problem 55 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} 0.4 x+0.2 y=8 \\ 0.7 x-0.3 y=1 \end{array}\right.$$

Step-by-Step Solution

Answer

x=10, y=20.

Step 1: Multiply to Clear Decimals

To clear the decimals, multiply each equation by 10. This gives: \[\begin{array}{l} 4x + 2y = 80 \ 7x - 3y = 10 \end{array}\]

Step 2: Use the Elimination Method

To eliminate one of the variables, multiply the first equation by 3 and the second equation by 2. This gives: \[\begin{array}{l} (3)(4x + 2y) = (3)(80) \ (2)(7x - 3y) = (2)(10) \end{array}\] Which simplifies to: \[\begin{array}{l} 12x + 6y = 240 \ 14x - 6y = 20 \end{array}\]

Step 3: Add the Equations

Add the two equations together to eliminate the variable y: \[12x + 6y + 14x - 6y = 240 + 20\] which simplifies to \[26x = 260\] Now solve for x: \[x = \frac{260}{26} = 10\]

Step 4: Substitute to Find y

Substitute x = 10 back into one of the original equations to find y. We use the first equation: \[4(10) + 2y = 80\] \[40 + 2y = 80\] Subtract 40 from both sides: \[2y = 40\] Divide by 2: \[y = \frac{40}{2} = 20\]

Step 5: Verify the Solution

Substitute x = 10 and y = 20 into the second original equation to verify: \[0.7(10) - 0.3(20) = 1\] \[7 - 6 = 1\] The solution is verified.

Key Concepts

elimination method

The elimination method is a powerful tool for solving systems of equations. The core idea is to eliminate one variable so you can focus on solving for the other. In our example, we started by writing two equations:

• 0.4x + 0.2y = 8
• 0.7x - 0.3y = 1

To make it easier to work with, we first eliminated the decimals by multiplying each equation by 10. Next, we lined up our equations in such a way that we could add or subtract them to eliminate one variable. After setting up our equations, we multiplied the first equation by 3 and the second by 2:

• 4x + 2y = 80 becomes 12x + 6y = 240
• 7x - 3y = 10 becomes 14x - 6y = 20

Adding these gives you a clear path to find one variable, simplifying the entire process.

multiply to clear decimals

Dealing with decimals can complicate calculations. A good trick here is to clear them by multiplying the entire equation by the appropriate factor, typically a power of 10. This simplifies each term, making your equations easier to handle. For our example, we had:

• 0.4x + 0.2y = 8
• 0.7x - 0.3y = 1

By multiplying everything by 10, we cleared the decimals and turned the equations into:

• 4x + 2y = 80
• 7x - 3y = 10

These are much cleaner and easier to work with. This step often makes subsequent operations like addition, subtraction, and even multiplication simpler and more straightforward.

substitute to find variable

Once you have isolated one of the variables using the elimination method, the next step is to substitute it back into one of the original equations. In our case, we found that x = 10. The substitution step involves replacing x in either equation to solve for y. Here’s how we did it:

• 4(10) + 2y = 80

When simplified, this becomes:

• 40 + 2y = 80

From there, we subtract 40 from both sides and find:

• 2y = 40

Dividing by 2 gives us y = 20. Substitution ensures we have consistent values for both variables in the system of equations.

verify solution

The final step in solving a system of equations is to verify your solution. This ensures that the variables you found actually satisfy the original equations. Substituting x = 10 and y = 20 back into the second original equation, we get:

• 0.7(10) - 0.3(20) = 1

Simplifying that gives:

• 7 - 6 = 1

Since both sides of the equation match, our solution is verified. This is an essential step to confirm that no mistakes were made during calculations and that the solution fits both equations perfectly.