Solve the system by elimination or by any convenient method. (If the system is dependent, enter DEPENDENT. If there is no solution, enter NO SOLUTION.)

*x*

−

1

4

*y*

=

1

4

5

*x*

−

1

5

*y*

=

1

We have the following system of linear equations:

\[
\begin{aligned}
1) & \quad x - \frac{1}{4}y = \frac{1}{4} \\
2) & \quad 5x - \frac{1}{5}y = 1
\end{aligned}
\]

Let's solve this system using the **elimination method**:

### Step 1: Eliminate \(y\)

We want to eliminate \(y\) by multiplying the two equations by appropriate factors so that the coefficients of \(y\) match.

To eliminate \(y\), we will multiply the first equation by 5 and the second equation by 4 to match the coefficients of \(y\).

\[
\begin{aligned}
5 \times \left( x - \frac{1}{4}y = \frac{1}{4} \right) & : \quad 5x - \frac{5}{4}y = \frac{5}{4} \\
4 \times \left( 5x - \frac{1}{5}y = 1 \right) & : \quad 20x - \frac{4}{5}y = 4
\end{aligned}
\]

The system becomes:
\[
\begin{aligned}
5x - \frac{5}{4}y &= \frac{5}{4} \\
20x - \frac{4}{5}y &= 4
\end{aligned}
\]

### Step 2: Eliminate \(y\) by subtracting equations

We need to multiply the equations by numbers that make the coefficients of \(y\) the same. To do this, we will multiply both equations by the least common multiple (LCM) of 4 and 5, which is 20.

\[
\begin{aligned}
20 \times \left( 5x - \frac{5}{4}y = \frac{5}{4} \right) & : \quad 100x - 25y = 25 \\
20 \times \left( 20x - \frac{4}{5}y = 4 \right) & : \quad 400x - 16y = 80
\end{aligned}
\]

The system now is:
\[
\begin{aligned}
100x - 25y &= 25 \\
400x - 16y &= 80
\end{aligned}
\]

Now, we subtract the first equation from the second:

\[
(400x - 16y) - (100x - 25y) = 80 - 25
\]

Simplifying:

\[
300x + 9y = 55
\]

This system still contains \(x\) and \(y\), but we may benefit from adjusting previous

Solve the system by elimination or by any convenient method. (If the system is dependent, enter DEPENDENT. If there is no solution, enter NO SOLUTION.)

x - 1/4 y = 1/4
5x - 1/5 y = 1

Learn how to solve the system of linear equations x - 1/4y = 1/4 and 5x - 1/5y = 1 using the elimination method. This problem covers key algebra concepts and is suitable for students in Grades 9-11.

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