Solve the following system by the elimination method.

one fifth x plus y
equalssix fifths
one tenth x plus one third y
equalsseven tenths

1 answer

To solve the system using the elimination method, let's first rewrite the equations based on your input:

  1. \(\frac{1}{5} x + y = \frac{6}{5}\)
  2. \(\frac{1}{10} x + \frac{1}{3} y = \frac{7}{10}\)

Next, we'll eliminate the fractions by multiplying both equations by the least common multiple (LCM) of the denominators.

Step 1: Eliminate Fractions

For the first equation, the LCM of 5 is 5. Multiply the entire equation by 5:

\[ 5\left(\frac{1}{5} x + y\right) = 5\left(\frac{6}{5}\right) \]

This simplifies to:

\[ x + 5y = 6 \quad \text{(Equation 1)} \]

For the second equation, the LCM of 10 and 3 is 30. Multiply the entire equation by 30:

\[ 30\left(\frac{1}{10} x + \frac{1}{3} y\right) = 30\left(\frac{7}{10}\right) \]

This simplifies to:

\[ 3x + 10y = 21 \quad \text{(Equation 2)} \]

Step 2: Set Up the New System

Now we have the following system of equations:

  1. \(x + 5y = 6\)
  2. \(3x + 10y = 21\)

Step 3: Elimination Method

We can multiply Equation 1 by 3 to eliminate \(x\):

\[ 3(x + 5y) = 3(6) \] which gives: \[ 3x + 15y = 18 \quad \text{(Equation 3)} \]

Now we have the modified system:

  1. \(3x + 15y = 18\) (from Equation 1)
  2. \(3x + 10y = 21\) (Equation 2)

Step 4: Subtract Equations

Subtract Equation 2 from Equation 3:

\[ (3x + 15y) - (3x + 10y) = 18 - 21 \]

This simplifies to:

\[ 5y = -3 \]

Step 5: Solve for \(y\)

Divide both sides by 5:

\[ y = -\frac{3}{5} \]

Step 6: Substitute \(y\) Back to Find \(x\)

Now substitute \(y\) back into one of the original equations. We'll use Equation 1:

\[ x + 5\left(-\frac{3}{5}\right) = 6 \]

This simplifies to:

\[ x - 3 = 6 \]

Adding 3 to both sides gives:

\[ x = 9 \]

Step 7: Final Solution

The solution to the system is:

\[ x = 9, \quad y = -\frac{3}{5} \]

Thus, the final answer is:

\[ (x, y) = \left(9, -\frac{3}{5}\right) \]