Question

Solve the following system of equations using the elimation method. 0.1x +0.3y = 0.9, 0.1x -0.3y =0.2

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

1. \( 0.1x + 0.3y = 0.9 \)
2. \( 0.1x - 0.3y = 0.2 \)

We can eliminate \( x \) by subtracting the second equation from the first:

\[
(0.1x + 0.3y) - (0.1x - 0.3y) = 0.9 - 0.2
\]

This simplifies to:

\[
0.1x + 0.3y - 0.1x + 0.3y = 0.7
\]

Combining like terms yields:

\[
0.6y = 0.7
\]

Now, we can solve for \( y \):

\[
y = \frac{0.7}{0.6} = \frac{7}{6} \approx 1.1667
\]

Next, we substitute \( y \) back into one of the original equations to solve for \( x \). We'll use the first equation:

\[
0.1x + 0.3\left(\frac{7}{6}\right) = 0.9
\]

Calculating \( 0.3 \times \frac{7}{6} \):

\[
0.3 \times \frac{7}{6} = \frac{2.1}{6} = \frac{7}{20}
\]

Therefore, we substitute this back into the first equation:

\[
0.1x + \frac{7}{20} = 0.9
\]

Now, we need to isolate \( x \):

\[
0.1x = 0.9 - \frac{7}{20}
\]

To perform this subtraction, convert 0.9 to a fraction with a denominator of 20:

\[
0.9 = \frac{18}{20}
\]

So:

\[
0.1x = \frac{18}{20} - \frac{7}{20} = \frac{11}{20}
\]

Dividing both sides by 0.1 (or multiplying by 10):

\[
x = \frac{11}{20} \times 10 = \frac{110}{20} = \frac{11}{2} = 5.5
\]

Thus, the solution to the system of equations is:

\[
x = 5.5, \quad y = \frac{7}{6} \approx 1.1667
\]

In conclusion, the solution is:

\[
\boxed{(5.5, \frac{7}{6})}
\]