To solve the given system of simultaneous equations:
1. [tex]\[\frac{x+1}{3}+\frac{y}{2}=0\][/tex]
2. [tex]\[\frac{x}{2}-2=\frac{y-1}{3}\][/tex]
we will go through a step-by-step procedure to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
### Step 1: Clear the Fractions
First, let's eliminate fractions by multiplying each equation by the least common multiple (LCM) of the denominators.
Equation 1:
[tex]\[
\frac{x+1}{3} + \frac{y}{2} = 0
\][/tex]
The LCM of 3 and 2 is 6. Multiply the entire equation by 6:
[tex]\[6 \left(\frac{x+1}{3}\right) + 6 \left(\frac{y}{2}\right) = 6 \cdot 0 \][/tex]
Simplifying:
[tex]\[2(x+1) + 3y = 0\][/tex]
[tex]\[2x + 2 + 3y = 0\][/tex]
Rewriting it:
[tex]\[2x + 3y = -2 \quad \text{(Equation 3)}\][/tex]
Equation 2:
[tex]\[
\frac{x}{2} - 2 = \frac{y-1}{3}
\][/tex]
The LCM of 2 and 3 is 6. Multiply the entire equation by 6:
[tex]\[6 \left(\frac{x}{2}\right) - 6 \cdot 2 = 6 \left(\frac{y-1}{3}\right)\][/tex]
Simplifying:
[tex]\[3x - 12 = 2(y - 1)\][/tex]
[tex]\[3x - 12 = 2y - 2\][/tex]
Rewriting it:
[tex]\[3x - 2y = 10 \quad \text{(Equation 4)}\][/tex]
### Step 2: Solve the System of Linear Equations
Now we have the system of equations:
1. [tex]\(2x + 3y = -2\)[/tex]
2. [tex]\(3x - 2y = 10\)[/tex]
We can use the method of elimination or substitution to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Let's use elimination:
Elimination Method:
To eliminate [tex]\(y\)[/tex], let's multiply Equation 3 by 2 and Equation 4 by 3 to make the coefficients of [tex]\(y\)[/tex] equal:
Multiply Equation 3 by 2:
[tex]\[4x + 6y = -4 \quad \text{(Equation 5)}\][/tex]
Multiply Equation 4 by 3:
[tex]\[9x - 6y = 30 \quad \text{(Equation 6)}\][/tex]
Add Equation 5 and Equation 6:
[tex]\[
(4x + 6y) + (9x - 6y) = -4 + 30
\][/tex]
[tex]\[
4x + 9x + 6y - 6y = 26
\][/tex]
[tex]\[
13x = 26
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{26}{13}
\][/tex]
[tex]\[
x = 2
\][/tex]
### Step 3: Substitute [tex]\(x\)[/tex] Back to Find [tex]\(y\)[/tex]
Now substitute [tex]\(x = 2\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We'll use Equation 3:
[tex]\[2(2) + 3y = -2\][/tex]
[tex]\[4 + 3y = -2\][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[3y = -2 - 4\][/tex]
[tex]\[3y = -6\][/tex]
[tex]\[y = \frac{-6}{3}\][/tex]
[tex]\[y = -2\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
x = 2, \quad y = -2
\][/tex]
