[tex]\large\underline{\sf{Solution-}}[/tex]
#1
Solving by submitting method:
[tex]\sf \leadsto 5x - 4y = 9 \: \: - - - (i)[/tex]
[tex]\sf \leadsto x - 2y = -3 \: \: - - - (ii)[/tex]
By first equation,
[tex]\sf \leadsto 5x - 4y = 9[/tex]
[tex]\sf \leadsto 5x = 9 + 4y[/tex]
[tex]\sf\leadsto x = \frac{9+4y}{5}\\[/tex]
Now, we can find the original value of y from Eqⁿ (ii).
[tex]\sf \leadsto x - 2y = -3[/tex]
[tex]\sf \leadsto \bigg(\frac{9+4y}{5}\bigg) - 2y = -3\\[/tex]
[tex]\sf \leadsto \frac{9+4y-10y}{5} = -3\\[/tex]
[tex]\sf \leadsto \frac{9-6y}{5}= -3\\[/tex]
[tex]\sf \leadsto 9-6y = -3(5)[/tex]
[tex]\sf \leadsto 9 -6y = - 15[/tex]
[tex]\sf \leadsto -6y = - 15 - 9[/tex]
[tex]\sf \leadsto -6y = -6[/tex]
[tex]\sf \leadsto y = \frac{\cancel-6}{\cancel-6}\\ [/tex]
[tex]\sf \leadsto y = \dfrac{ 6}{ 6} \\[/tex]
[tex]\sf \leadsto y = 1[/tex]
Now, we can find the original value of x , from Eqⁿ (I)
[tex]\sf \leadsto 5x-4y = 9[/tex]
[tex]\sf \leadsto 5x - 4(1) = 9[/tex]
[tex]\sf \leadsto 5x - 1 = 9[/tex]
[tex]\sf \leadsto 5x = 9+1[/tex]
[tex]\sf \leadsto 5x = 10[/tex]
[tex]\sf \leadsto x = \frac{10}{5}\\[/tex]
[tex]\sf \leadsto x = 5[/tex]
Therefore, the values of x and y are 5 and 1 respectively.
#2
Solving by Eliminating method:
[tex]\sf \leadsto 4x + 6y = 12\:\:- - - (i)[/tex]
[tex]\sf \leadsto 6x+9y = 12 \:\: - - -(ii)[/tex]
Step 1: Multiply Eqⁿ (I) by 3 and Eqⁿ (ii) by 2 to make the coefficients of y equal. Then we get the equations.
[tex]\sf \leadsto 12x + 18y = 36\:\: - - - (iii) [/tex]
[tex]\sf \leadsto 12x + 18y = 24\:\:- - - (iv)[/tex]
Step 2: Subtract Eqⁿ (iii) from Eqⁿ (iv) , we get
[tex]\sf \leadsto (12x + 18y)-(12x + 18y) = 36-24[/tex]
[tex]\sf \leadsto 0 =12, Which is a false statement. [/tex]
Therefore, the pair of equations has no solution.
Hope this helps!!
