1. Solve by elimination and substitution

(i) $x+y=5,2x-3y=4$

Elimination:

• Multiply $x+y=5$ by 2: $2x+2y=10$
  

• Subtract $2x-3y=4$ from this: $(2x+2y)-(2x-3y)=10-4\Rightarrow 5y=6$
  

• So $y={\displaystyle \frac{6}{5}}$. Then $x=5-{\displaystyle \frac{6}{5}}={\displaystyle \frac{19}{5}}$.

Substitution (quick):

• $x=5-y$. Put in $2x-3y=4$
  $2(5-y)-3y=4\Rightarrow 10-5y=4\Rightarrow y=6\mathrm{/}5$, same result.

Answer: $\boxed{{\displaystyle (x,y)=(\frac{19}{5},\frac{6}{5})}}\mathrm{}$

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(ii) $3x+4y=10,2x-2y=2$

Elimination:

• Multiply second eqn by 2: $4x-4y=4$
  

• Add to first: $(3x+4y)+(4x-4y)=10+4\Rightarrow 7x=14\Rightarrow x=2$
  

• Then $3(2)+4y=10\Rightarrow 6+4y=10\Rightarrow y=1$
  

Substitution would give same.

Answer: $\boxed{{\displaystyle (x,y)=(2,1)}}\mathrm{}$

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(iii) $3x-5y-4=0$ and $9x=2y+7.$
Rewrite:

• $3x-5y=4$ and $9x-2y=7$

Elimination:

• Multiply first by 3: $9x-15y=12$
  

• Subtract the second: $(9x-15y)-(9x-2y)=12-7\Rightarrow -13y=5\Rightarrow y=-{\displaystyle \frac{5}{13}}$
  

• Then $3x-5(-5\mathrm{/}13)=4\Rightarrow 3x+{\displaystyle \frac{25}{13}}=4\Rightarrow 3x={\displaystyle \frac{27}{13}}\Rightarrow x={\displaystyle \frac{9}{13}}$
  

Answer: $\boxed{{\displaystyle (x,y)=(\frac{9}{13},-\frac{5}{13})}}\mathrm{}$

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(iv) ${\displaystyle \frac{x}{2}}+{\displaystyle \frac{2y}{3}}=-1,x-{\displaystyle \frac{y}{3}}=3$

First clear denominators:

• Multiply first by 6: $3x+4y=-6$
  

• Multiply second by 3: $3x-y=9$

Elimination:

• Subtract the second from the first: $(3x+4y)-(3x-y)=-6-9\Rightarrow 5y=-15\Rightarrow y=-3.$
  

• Then $3x-(-3)=9\Rightarrow 3x+3=9\Rightarrow 3x=6\Rightarrow x=2.$
  

Answer: $\boxed{{\displaystyle (x,y)=(2,-3)}}\mathrm{}$

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2. Word problems — form the pair and solve (elimination)

(i) Fraction problem.
Let fraction $={\displaystyle \frac{x}{y}}$. Conditions:

$$\begin{gathered} \frac{x+1}{y-1}=1\Rightarrow x+1=y-1\Rightarrow x-y=-2. \\ \frac{x}{y+1}=\frac{1}{2}\Rightarrow 2x=y+1. \end{gathered}$$

Solve: from $x-y=-2$ ⇒ $x=y-2$

Put in $2x=y+1$: $2(y-2)=y+1\Rightarrow 2y-4=y+1\Rightarrow y=5$.

Then $x=3$.

Answer: The fraction is $\boxed{{\displaystyle \frac{3}{5}}}\mathrm{}$

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(ii) Ages problem (Nuri & Sonu).
Let Nuri $=N$, Sonu $=S$

$$\text{Five years ago: }N-5=3(S-5)\Rightarrow N-3S=-10.$$

$$\text{Ten years later: }N+10=2(S+10)\Rightarrow N-2S=10.$$

Subtract first from second: $(N-2S)-(N-3S)=10-(-10)\Rightarrow S=20.$ Then $N-2(20)=10\Rightarrow N=50.$

Answer: Nuri $=\mathrm{50 }$years, Sonu $=\mathrm{20 }$years.

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(iii) Two-digit number.
Let tens digit $=x$, units $=y$. Then number $=10x+y$
Given $x+y=9$.
Also $9(10x+y)=2(10y+x)$
Compute: $90x+9y=20y+2x\Rightarrow 88x=11y\Rightarrow 8x=y$.
Combine with $x+y=9$
$x+8x=9\Rightarrow 9x=9\Rightarrow x=1,y=8$
Number $=18$

Answer: $\boxed{{\displaystyle 18}}\mathrm{.}$