Solve using elimination method: 6x - 5y = 0; 2x - y = 4.
Understand the Problem
The question is asking us to solve a system of linear equations using the elimination method. This involves eliminating one variable by combining the two equations in such a way that we can solve for the remaining variable, then substituting back to find the other variable.
Answer
The solution is $x = \frac{3}{2}$ and $y = 1$.
Answer for screen readers
The solution to the system of equations is $x = \frac{3}{2}$ and $y = 1$.
Steps to Solve
-
Write down the equations
Identify the given system of linear equations. For example:
$$
2x + 3y = 6 \quad (1)
4x - y = 5 \quad (2)
$$ -
Choose a variable to eliminate
Decide which variable you want to eliminate. Let's eliminate $y$ here. We can do this by modifying the equations so that the coefficients of $y$ in both equations are equal. -
Multiply the equations if necessary
Multiply equation (2) by $3$ to align the coefficients of $y$:
$$
3(4x - y) = 3(5) \implies 12x - 3y = 15 \quad (3)
$$
So the new equations are:
$$
2x + 3y = 6 \quad (1)
12x - 3y = 15 \quad (3)
$$ -
Add the equations
Now, add equations (1) and (3) together to eliminate $y$:
$$
(2x + 3y) + (12x - 3y) = 6 + 15
$$
This simplifies to:
$$
14x = 21
$$ -
Solve for the remaining variable
Now, divide both sides by $14$ to find $x$:
$$
x = \frac{21}{14} = \frac{3}{2}
$$ -
Substitute back to find the other variable
Substitute $x$ back into one of the original equations to find $y$. We use equation (1):
$$
2\left(\frac{3}{2}\right) + 3y = 6
$$
This simplifies to:
$$
3 + 3y = 6
$$
Now, solve for $y$:
$$
3y = 6 - 3
$$
$$
3y = 3 \implies y = 1
$$ -
Write the final solution
The solution of the system is:
$$
x = \frac{3}{2}, \quad y = 1
$$
The solution to the system of equations is $x = \frac{3}{2}$ and $y = 1$.
More Information
In this system of equations, one variable was successfully eliminated using the elimination method, allowing us to easily solve for the other variable.
Tips
- Failing to multiply the equations correctly to match coefficients.
- Forgetting to substitute back the found value into the original equations.
- Miscalculating simple arithmetic when solving for variables.