fate. 10/0.3120251 Term1 Day. Monday week 7 Assignment \# 1: Solving simultaneous Equations 1 Solue for $$ x $$ and $$ Y $$ : $$ \begin{array}{l} 3 x+2 y=12 \ x+2 y=8 \end{array} $$ 2 Solve by elimination. $$ \begin{array}{l} 2 x-y=4 \ 3 x+y=11 \end{array} $$ 3 Solve using substitution: $$ \begin{array}{l} y=2 x+1 \ y=x+4 \end{array} $$ 4 Solve using multiplication and addition $$ \begin{array}{l} 2 x+3 y=12 \ 4 x-y=5 \end{array} $$ 5 Solve the equation with parentheses: $$ \begin{array}{l} 2(x+3)+y=10 \ 3 x-y=5 \end{array} $$ 6 Solve for $$ x $$ and $$ y $$. $$ \begin{array}{l} x+y=10 \ 2 x-y=4 \end{array} $$ 7 Solve the froctional equation. $$ \begin{array}{l} \frac{3 x}{2}+y=7 \ x-\frac{y}{3}=2 \end{array} $$

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1. We are given:
   $$
   \begin{aligned}
   3x+2y &= 12 \
   x+2y &= 8
   \end{aligned}
   $$
   - Subtract the second equation from the first:
     $$
     (3x+2y)-(x+2y)=12-8 \quad\Longrightarrow\quad 2x=4
     $$
     $$
     \Longrightarrow\quad x=2
     $$
   - Substitute $$x=2$$ into $$x+2y=8$$:
     $$
     2+2y=8 \quad\Longrightarrow\quad 2y=6
     $$
     $$
     \Longrightarrow\quad y=3
     $$

2. We have:
   $$
   \begin{aligned}
   2x-y &= 4 \
   3x+y &= 11
   \end{aligned}
   $$
   - Add both equations to eliminate $$y$$:
     $$
     (2x-y)+(3x+y)=4+11 \quad\Longrightarrow\quad 5x=15
     $$
     $$
     \Longrightarrow\quad x=3
     $$
   - Substitute $$x=3$$ into $$3x+y=11$$:
     $$
     3(3)+y=11 \quad\Longrightarrow\quad 9+y=11
     $$
     $$
     \Longrightarrow\quad y=2
     $$

3. The system is:
   $$
   \begin{aligned}
   y &= 2x+1 \
   y &= x+4
   \end{aligned}
   $$
   - Set the expressions for $$y$$ equal:
     $$
     2x+1=x+4 \quad\Longrightarrow\quad 2x-x=4-1
     $$
     $$
     \Longrightarrow\quad x=3
     $$
   - Substitute $$x=3$$ into $$y=x+4$$:
     $$
     y=3+4=7
     $$

4. The system is:
   $$
   \begin{aligned}
   2x+3y &= 12 \
   4x-y &= 5
   \end{aligned}
   $$
   - Multiply the second equation by 3:
     $$
     3(4x-y)=3(5) \quad\Longrightarrow\quad 12x-3y=15
     $$
   - Add the first equation:
     $$
     (2x+3y)+(12x-3y)=12+15 \quad\Longrightarrow\quad 14x=27
     $$
     $$
     \Longrightarrow\quad x=\frac{27}{14}
     $$
   - Substitute $$x=\frac{27}{14}$$ into $$4x-y=5$$:
     $$
     4\left(\frac{27}{14}\right)-y=5 \quad\Longrightarrow\quad \frac{108}{14}-y=5
     $$
     $$
     \Longrightarrow\quad y=\frac{108}{14}-5=\frac{108-70}{14}=\frac{38}{14}=\frac{19}{7}
     $$

5. The equations are:
   $$
   \begin{aligned}
   2(x+3)+y &= 10 \
   3x-y &= 5
   \end{aligned}
   $$
   - Expand the first equation:
     $$
     2x+6+y=10 \quad\Longrightarrow\quad 2x+y=4
     $$
   - Add the equations:
     $$
     (2x+y)+(3x-y)=4+5 \quad\Longrightarrow\quad 5x=9
     $$
     $$
     \Longrightarrow\quad x=\frac{9}{5}
     $$
   - Substitute $$x=\frac{9}{5}$$ into $$2x+y=4$$:
     $$
     2\left(\frac{9}{5}\right)+y=4 \quad\Longrightarrow\quad \frac{18}{5}+y=4
     $$
     $$
     \Longrightarrow\quad y=4-\frac{18}{5}=\frac{20-18}{5}=\frac{2}{5}
     $$

6. The system is:
   $$
   \begin{aligned}
   x+y &= 10 \
   2x-y &= 4
   \end{aligned}
   $$
   - Solve the first equation for $$y$$:
     $$
     y=10-x
     $$
   - Substitute into the second equation:
     $$
     2x-(10-x)=4 \quad\Longrightarrow\quad 2x-10+x=4
     $$
     $$
     \Longrightarrow\quad 3x=14 \quad\Longrightarrow\quad x=\frac{14}{3}
     $$
   - Substitute $$x=\frac{14}{3}$$ into $$y=10-x$$:
     $$
     y=10-\frac{14}{3}=\frac{30-14}{3}=\frac{16}{3}
     $$

7. The system is:
   $$
   \begin{aligned}
   \frac{3x}{2}+y &= 7 \
   x-\frac{y}{3} &= 2
   \end{aligned}
   $$
   - Multiply the first equation by 2 to eliminate the fraction:
     $$
     3x+2y=14
     $$
   - Multiply the second equation by 3:
     $$
     3x-y=6
     $$
   - Subtract the second new equation from the first:
     $$
     (3x+2y)-(3x-y)=14-6 \quad\Longrightarrow\quad 3y=8
     $$
     $$
     \Longrightarrow\quad y=\frac{8}{3}
     $$
   - Substitute $$y=\frac{8}{3}$$ into $$3x-y=6$$:
     $$
     3x-\frac{8}{3}=6 \quad\Longrightarrow\quad 3x=6+\frac{8}{3}=\frac{18+8}{3}=\frac{26}{3}
     $$
     $$
     \Longrightarrow\quad x=\frac{26}{9}
     $$
1. $$ x = 2 $$, $$ y = 3 $$ 2. $$ x = 3 $$, $$ y = 2 $$ 3. $$ x = 3 $$, $$ y = 7 $$ 4. $$ x = \frac{27}{14} $$, $$ y = \frac{19}{7} $$ 5. $$ x = \frac{9}{5} $$, $$ y = \frac{2}{5} $$ 6. $$ x = \frac{14}{3} $$, $$ y = \frac{16}{3} $$ 7. $$ x = \frac{26}{9} $$, $$ y = \frac{8}{3} $$

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