Step-by-step explanation:

2x+y-5=0

2x+y=5

y=5-2x ......(i)

3x+2y-8=0

3x+2y=8 .......(ii)

substituting eq. (i) in (ii)

3x+2(5-2x)=8

3x+10-4x=8

10-x=8

-x=8-10

-x=-2

x=2 ......(iii)

substituting eq. (iii) in (i)

y=5-2(2)

y=5-4

y=1 ......(iv)

Answer:

[tex] \large \green{\boxed{ \bold{x = 2}}}[/tex]

[tex] \large \green{\boxed{ \bold{y = 1}}}[/tex]

Step-by-step explanation:

Given pair of linear equations :

[tex] \large2x + y - 5 = 0[/tex]

[tex] \large3x + 2y - 8 = 0[/tex]

Let us consider the first linear equation,

[tex] \large2x + y - 5 = 0[/tex]

[tex] \large \implies 2x + y = 5[/tex]

[tex] \large \implies y = 5 - 2x ................(i)[/tex]

Now, let us consider the second linear equation,

[tex] \large3x + 2y - 8 = 0[/tex]

[tex] \large \implies 3x + 2y = 8[/tex]

[tex] \large \implies3x = 8 - 2y[/tex]

[tex] \large \implies x = \dfrac{8 - 2y}{3}..............(ii)[/tex]

Then, on substituting equation (ii) in equation (i), we get:

[tex] \large y = 5 - 2x[/tex]

[tex] \large \implies y = 5 - 2 \times \bigg ( \dfrac{8 - 2y}{3} \bigg)[/tex]

[tex] \large \implies y = 5 - \dfrac{16 + 4y}{3} [/tex]

[tex] \large \implies y = \dfrac{15 - 16 + 4y}{3} [/tex]

[tex] \large \implies y = \dfrac{ - 1 + 4y}{3} [/tex]

[tex] \large \implies3 y = - 1 + 4y [/tex]

[tex] \large \implies3 y - 4y = - 1[/tex]

[tex] \large \implies - y = - 1[/tex]

[tex] \large \boxed{ \bold{\therefore y = 1}}[/tex]

At last, substituting (y = 1) in equation (ii), we get:

[tex] \large \implies x = \dfrac{8 - 2y}{3}[/tex]

[tex] \large \implies x = \dfrac{8 - (2 \times 1)}{3}[/tex]

[tex] \large \implies3 x = 8 - 2 [/tex]

[tex] \large \implies 3x = 6[/tex]

[tex] \large \implies x = \dfrac{ \cancel6}{ \cancel3}[/tex]

[tex] \large \boxed{ \bold{\therefore x = 2}}[/tex]

[tex] \red{ \rule{190pt}{5pt}}[/tex]