solve using cross multiplication method:
2x=5y+4
3x-2y+16=0

Answer :

We know if we have equation :  a₁ x  +  b₁ y + c₁ = 0   and    a₂ x  +  b₂ y + c₂ = 0 

So, from cross multiplication method we know :

x  = $\frac{\left( {\mathrm{b}}_1{\mathrm{c}}_2 - {\mathrm{b}}_2{\mathrm{c}}_1 \right)}{\left( {\mathrm{a}}_1{\mathrm{b}}_2 - {\mathrm{a}}_2{\mathrm{b}}_1 \right)}$ and y  = $\frac{\left( {\mathrm{c}}_1{\mathrm{a}}_2 - {\mathrm{c}}_2{\mathrm{a}}_1 \right)}{\left( {\mathrm{a}}_1{\mathrm{b}}_2 - {\mathrm{a}}_2{\mathrm{b}}_1 \right)}$

Here we have equations : 2 x  - 5  y  - 4   = 0   and  3 x  - 2 y + 16  = 0  

So,

a₁  = 2  ,  b₁ = - 5 , c₁ = - 4   and    a₂  = 3,  b₂   = - 2 , c₂ = 16

Then ,

x  = $\frac{\left(\left(- 5\right)\times 16 - \left(- 2\right)\times \left(- 4\right) \right)}{\left( 2\times \left(- 2\right) -3 \times \left(- 5\right) \right)} = \frac{\left( - 80 -8 \right)}{\left( - 4 +\hspace{0.17em}15 \right)}= \frac{ - 88 }{ 11}= \mathbf{-}\mathbf{ }\mathbf{8}$

And

y  = $\frac{\left(\left(- 4\right)\times 3 - 16\times 2 \right)}{\left( 2\times \left(- 2\right) -3 \times \left(- 5\right) \right)} = \frac{\left( - 12 -32 \right)}{\left( - 4 +\hspace{0.17em}15 \right)}= \frac{ - 44 }{ 11}= \mathbf{-}\mathbf{ }\mathbf{4}$

So,

x  = - 8   and y  = - 4                                                                        ( Ans )