Question

Solve the following system by the elimination method.
one fifth x plus y
equalssix fifths
one tenth x plus one third y
equalsseven tenths

Answer 1

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:

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