Math Problem Statement

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

Solution

We have the following system of linear equations:

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

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations
Elimination Method

Formulas

Elimination Method for Systems of Equations

Theorems

No specific theorem, but involves linear equation solving techniques

Suitable Grade Level

Grades 9-11

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