Question 59808
(1) {{{x/3+4y=30}}}
(2) {{{2x-y=5}}}

Add y to both sides of equation (2) to get {{{2x=5+y}}}
Then subtract 5 from both sides and you get {{{y=2x-5}}}
Replace y with {{{2x-5}}} in (1) and you get 

(3) {{{x/3+4(2x-5)=30}}}

Multiply 4 by 2x-5 to get 8x-20 so (3) becomes

{{{x/3+8x-20=30}}}

Add 20 to both sides to get 

(4) {{{x/3+8x=50}}}

Add {{{x/3}}} to {{{8x}}} to get {{{(8+1/3)x}}}
But, {{{8=24/3}}} so {{{(8+1/3)x=(24/3+1/3)x=(25/3)x}}}

So, replacing {{{x/3+8x}}} with {{{(25/3)x}}} in (4) we get
{{{(25/3)x=50}}}

Now, multiply both sides by {{{3/25}}} and we get {{{x=50(3/25)}}} or x=6.
Since x=6, we replace x with 6 in (2) and get {{{2*6-y=5}}} or {{{12-y=5}}} so y=7.

To verify that x=6, y=7 is correct we use these values in (1) and (2):

From (1): {{{x/3+4y=30}}} or {{{6/3+4*7=2+28}}} which is indeed 30, and
From (2): {{{2x-y=5}}} or {{{2*6-7=12-7}}} which is indeed 5.