Answer:
Step-by-step explanation:
For Case -1
[tex]\Rightarrow -4x+3y=-5\quad \ldots(i)[/tex]
[tex]\Rightarrow 4x-5y=3[/tex]
[tex]\Rightarrow 4x=5y+3[/tex]
Put the value of [tex]4x[/tex] in equation [tex](i)[/tex] we get
[tex]\Rightarrow -(5y+3)+3y=-5[/tex]
[tex]\Rightarrow -5y+3y-3=-5[/tex]
[tex]\Rightarrow 2=2y[/tex]
[tex]\Rightarrow y=1[/tex]
and thus [tex]4x=5(1)+3[/tex]
[tex]x=\frac{8}{4}=2[/tex]
Case-2
[tex]10x-6y=12\quad \ldots(i)[/tex]
[tex]2x+3y=12\quad \ldots(ii)[/tex]
Now multiply [tex](ii) by 5 and subtract it from (i)[/tex]
[tex] 10x-6y=12[/tex]
[tex] 10x+15y=+180[/tex]
[tex]\cdots \cdots \cdots[/tex]
[tex]-21y=-168[/tex]
[tex]y=8[/tex]
and [tex]=6[/tex]
