Solve the pairs of simultaneous equatic a) $$ x+y=5 ; \quad x-y=3 $$ b) $$ x-3 y=-5 ; \quad 2 x+3 y=-1 $$ c) $$ 2 y-3 x=-21 ; \quad 2 x+3 y=-1 $$ d) $$ 3 x+2 y=60 ; \quad 3 x-3 y=45 $$ e) $$ x+y=36 ; \quad x-2 y=-12 $$ f) $$ x+2 y=5 ; \quad 3 x-y $$

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**a)**  
We have the system  
$$
x+y=5 \quad (1)
$$  
$$
x-y=3 \quad (2)
$$  
Add equations (1) and (2):  
$$
(x+y)+(x-y)=5+3 \quad\Longrightarrow\quad 2x=8
$$  
Thus,  
$$
x=4.
$$  
Substitute $$ x=4 $$ into (1):  
$$
4+y=5 \quad\Longrightarrow\quad y=1.
$$

---

**b)**  
The system is  
$$
x-3y=-5 \quad (1)
$$  
$$
2x+3y=-1 \quad (2)
$$  
Add equations (1) and (2):  
$$
(x-3y)+(2x+3y)= -5 + (-1) \quad\Longrightarrow\quad 3x=-6
$$  
Thus,  
$$
x=-2.
$$  
Substitute $$ x=-2 $$ into (1):  
$$
-2-3y=-5 \quad\Longrightarrow\quad -3y=-3 \quad\Longrightarrow\quad y=1.
$$

---

**c)**  
Given the system  
$$
2y-3x=-21 \quad (1)
$$  
$$
2x+3y=-1 \quad (2)
$$  
Solve (2) for $$ x $$:  
$$
2x=-1-3y \quad\Longrightarrow\quad x=-\frac{1+3y}{2}.
$$  
Substitute this expression for $$ x $$ into (1):  
$$
2y-3\left(-\frac{1+3y}{2}\right)=-21.
$$  
Simplify:  
$$
2y+\frac{3(1+3y)}{2}=-21.
$$  
Multiply the entire equation by 2 to eliminate the fraction:  
$$
4y+3(1+3y)=-42.
$$  
Expand and combine like terms:  
$$
4y+3+9y=-42 \quad\Longrightarrow\quad 13y+3=-42.
$$  
Subtract 3 from both sides:  
$$
13y=-45 \quad\Longrightarrow\quad y=-\frac{45}{13}.
$$  
Now, substitute $$ y=-\frac{45}{13} $$ into the expression for $$ x $$:  
$$
x=-\frac{1+3\left(-\frac{45}{13}\right)}{2} 
=-\frac{1-\frac{135}{13}}{2}.
$$  
Write $$ 1 $$ as $$ \frac{13}{13} $$:  
$$
x=-\frac{\frac{13-135}{13}}{2}=-\frac{-122}{26}=\frac{122}{26}=\frac{61}{13}.
$$

---

**d)**  
The system is  
$$
3x+2y=60 \quad (1)
$$  
$$
3x-3y=45 \quad (2)
$$  
Subtract equation (2) from (1):  
$$
(3x+2y)-(3x-3y)=60-45 \quad\Longrightarrow\quad 5y=15.
$$  
Thus,  
$$
y=3.
$$  
Substitute $$ y=3 $$ into (1):  
$$
3x+2(3)=60 \quad\Longrightarrow\quad 3x+6=60.
$$  
Subtract 6:  
$$
3x=54 \quad\Longrightarrow\quad x=18.
$$

---

**e)**  
We are given  
$$
x+y=36 \quad (1)
$$  
$$
x-2y=-12 \quad (2)
$$  
Solve (1) for $$ x $$:  
$$
x=36-y.
$$  
Substitute into (2):  
$$
(36-y)-2y=-12 \quad\Longrightarrow\quad 36-3y=-12.
$$  
Subtract 36:  
$$
-3y=-48 \quad\Longrightarrow\quad y=16.
$$  
Then,  
$$
x=36-16=20.
$$

---

**f)**  
The system provided is  
$$
x+2y=5
$$  
$$
3x-y
$$  
The second equation is incomplete because it does not include an equality. Without a complete equation, we cannot solve the system. Please provide the full equation for part (f).
**a)** $$ x = 4 $$, $$ y = 1 $$ **b)** $$ x = -2 $$, $$ y = 1 $$ **c)** $$ x = \frac{61}{13} $$, $$ y = -\frac{45}{13} $$ **d)** $$ x = 18 $$, $$ y = 3 $$ **e)** $$ x = 20 $$, $$ y = 16 $$ **f)** Cannot solve due to incomplete information.

Solve the pairs of simultaneous equatic a) \( x+y=5 ; \quad x-y=3 \) b) \( x-3 y=-5 ; \quad 2 x+3 y=-1 \) c) \( 2 y-3 x=-21 ; \quad 2 x+3 y=-1 \) d) \( 3 x+2 y=60 ; \quad 3 x-3 y=45 \) e) \( x+y=36 ; \quad x-2 y=-12 \) f) \( x+2 y=5 ; \quad 3 x-y \)

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