To solve the system of linear equations:
[tex]\[ \frac{1}{3} x - \frac{1}{4} y = 1 \][/tex]
[tex]\[ \frac{2}{3} x - \frac{1}{2} y = 1 \][/tex]
we can use standard algebraic methods such as substitution or elimination. Here, we will use the elimination method to find values for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
First, we will eliminate the fractions to make the equations easier to work with. Let's multiply each term of the first equation by 12, the least common multiple of 3 and 4:
[tex]\[ 12 \left(\frac{1}{3} x - \frac{1}{4} y\right) = 12 \cdot 1 \][/tex]
[tex]\[ 4x - 3y = 12 \quad \text{(Equation 1)} \][/tex]
Now, let's multiply each term of the second equation by 6, the least common multiple of 3 and 2:
[tex]\[ 6 \left(\frac{2}{3} x - \frac{1}{2} y\right) = 6 \cdot 1 \][/tex]
[tex]\[ 4x - 3y = 6 \quad \text{(Equation 2)} \][/tex]
At this point, our system of equations is:
[tex]\[ 4x - 3y = 12 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 4x - 3y = 6 \quad \text{(Equation 2)} \][/tex]
We notice that the left-hand sides of both equations are identical, but the right-hand sides are different. This implies that the system of equations is inconsistent, meaning there are no values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that can simultaneously satisfy both equations.
In other words, there is no solution to this system of equations because the two lines represented by these equations are parallel and do not intersect.
Hence, the system has no solution.
