**Question 1**:

Solve the following simultaneous equations.

$\begin{array}{l}y+2x=2\\ \frac{2}{x}+\frac{1}{y}=5\end{array}$

**Solution**:$\begin{array}{l}y+2x=2----(1)\\ \frac{2}{x}+\frac{1}{y}=5----(2)\\ y=2-2x----(3)\\ \\ \text{substitute(3)into(2),}\\ \frac{2}{x}+\frac{1}{2-2x}=5\\ \frac{2\left(2-2x\right)+x}{x\left(2-2x\right)}=5\\ 4-4x+x=5x\left(2-2x\right)\\ 4-3x=10x-10{x}^{2}\\ 10{x}^{2}-13x+4=0\\ \left(5x-4\right)\left(2x-1\right)=0\\ 5x-4=0\text{or2}x-1=0\\ x=\frac{4}{5}\text{or}x=\frac{1}{2}\\ \\ \text{Substitutevaluesof}x\text{into(3),}\\ \text{When}x=\frac{4}{5}\text{},\text{}\\ y=2-2\left(\frac{4}{5}\right)=\frac{2}{5}\\ \\ \text{When}x=\frac{1}{2}\\ y=2-2(\frac{1}{2})=1\\ \\ \text{Thesolutionsare}x=\frac{4}{5},\text{}y=\frac{2}{5}\text{and}x=\frac{1}{2},\text{}y=1\end{array}$

**Question 2**:

Solve the following simultaneous equations.

*x*– 3

*y*+ 5 = 3

*y*+ 5

*y*

^{2}– 6 –

*x*= 0

*Solution:**x*– 3

*y*+ 5 = 0

*x*= 3

*y*– 5 —–(1)

3

*y*+ 5*y*^{2}– 6 –*x*= 0 —–(2)Substitute (1) into (2),

3

*y*+ 5*y*^{2}– 6 – (3*y*– 5) = 03

*y*+ 5*y*^{2}– 6 – 3*y*+ 5 = 05

*y*^{2}– 1 = 05

*y*^{2}= 1*y*= ±0.447

Substitute the values of

*y*into (1),When

*y*= 0.447*x*= 3 (0.447) – 5

*x*= –3.659

When

*y*= – 0.447*x*= 3 (–0.447) – 5

*x*= –6.341

**The solutions are**

*x*= –3.659,*y*= 0.447 and*x*= –6.341,*y*= – 0.447.
This site actually good but I want to suggest to provide more questions on this site such as exercise. Well, many student can learn if you provide more question . Thank you…..

2(2-x)+x = 4-3x not 4-4x

Thanks for pointing our mistake.

We have done the correction accordingly.