To solve the simultaneous equations:
[tex]\[
\frac{3}{x} - \frac{4}{y} = \frac{1}{3}
\][/tex]
[tex]\[
\frac{2}{x} - \frac{5}{y} = 1
\][/tex]
we will perform a series of algebraic manipulations to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
First, let us rewrite the equations by introducing substitution variables. Let [tex]\(u = \frac{1}{x}\)[/tex] and [tex]\(v = \frac{1}{y}\)[/tex]. Then, the equations become:
[tex]\[
3u - 4v = \frac{1}{3}
\][/tex]
[tex]\[
2u - 5v = 1
\][/tex]
Now we have a system of linear equations in terms of [tex]\(u\)[/tex] and [tex]\(v\)[/tex], which we can solve using methods such as substitution or elimination. We will use the elimination method to solve these equations.
### Step 1: Align the equations
[tex]\[
3u - 4v = \frac{1}{3} \quad \text{(equation 1)}
\][/tex]
[tex]\[
2u - 5v = 1 \quad \text{(equation 2)}
\][/tex]
### Step 2: Eliminate one variable
To eliminate one of the variables, we can multiply both equations by numbers that will allow the coefficients of [tex]\(u\)[/tex] or [tex]\(v\)[/tex] to cancel out when added or subtracted. We will multiply equation 1 by 2 and equation 2 by 3, so that the coefficients of [tex]\(u\)[/tex] become the same:
[tex]\[
2 \cdot (3u - 4v) = 2 \cdot \frac{1}{3} \implies 6u - 8v = \frac{2}{3} \quad \text{(equation 3)}
\][/tex]
[tex]\[
3 \cdot (2u - 5v) = 3 \cdot 1 \implies 6u - 15v = 3 \quad \text{(equation 4)}
\][/tex]
### Step 3: Subtract the rearranged equations to eliminate [tex]\(u\)[/tex]
[tex]\[
6u - 8v = \frac{2}{3}
\][/tex]
[tex]\[
6u - 15v = 3
\][/tex]
Subtract equation 3 from equation 4:
[tex]\[
(6u - 15v) - (6u - 8v) = 3 - \frac{2}{3}
\][/tex]
[tex]\[
6u - 15v - 6u + 8v = 3 - \frac{2}{3}
\][/tex]
[tex]\[
-7v = \frac{9}{3} - \frac{2}{3}
\][/tex]
[tex]\[
-7v = \frac{7}{3}
\][/tex]
Solving for [tex]\(v\)[/tex]:
[tex]\[
v = -\frac{1}{3}
\][/tex]
### Step 4: Substitute [tex]\(v\)[/tex] back to find [tex]\(u\)[/tex]
Substitute [tex]\(v = -\frac{1}{3}\)[/tex] into equation 1:
[tex]\[
3u - 4\left(-\frac{1}{3}\right) = \frac{1}{3}
\][/tex]
[tex]\[
3u + \frac{4}{3} = \frac{1}{3}
\][/tex]
Multiply all terms by 3 to clear the fraction:
[tex]\[
9u + 4 = 1
\][/tex]
[tex]\[
9u = -3
\][/tex]
[tex]\[
u = -\frac{1}{3}
\][/tex]
### Step 5: Substitute back to find [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
Remember, [tex]\(u = \frac{1}{x}\)[/tex] and [tex]\(v = \frac{1}{y}\)[/tex]:
[tex]\[
u = -\frac{1}{3} \implies \frac{1}{x} = -\frac{1}{3} \implies x = -3
\][/tex]
[tex]\[
v = -\frac{1}{3} \implies \frac{1}{y} = -\frac{1}{3} \implies y = -3
\][/tex]
So, the solution to the system of simultaneous equations is:
[tex]\[
x = -3, \quad y = -3
\][/tex]
