Question 76284
Once you have 2 simultaneous equations, you can solve for a variable in one equation by using a modified version of the other.

For this particular question, your equations are correct. So we have 
3y+5x=229
4y-25=3x

I'm going to start by putting the 2nd equation in terms of x, so when I solve for x I get:
4y-25=3x
{{{(4y-25)/3}}}=x (I divided each side by 3)

Now I can plug in that x value into the 1st equation, so I can solve for y.
3y+5x=229
3y+5*{{{((4y-25))/3}}}=229....distribute the 5 to everything in the parenthesis to get
3y+{{{(20y-125)/3}}}=229....now convert each term on the left side into like fractions of denominator 3 to get
{{{(9y/3)}}}+{{{(20y-125)/3}}}=229....now multiply each side by 3 to get rid of the fraction to get
9y+20y-125=687...add like terms to get
29y-125=687...add 125 to each side to get
29y=812...finally, we can divide to get
y=28

Back to the 2nd equation, we can now plug in a number for y, and then solve for x:
4y-25=3x
4(28)-25=3x
112-25=3x
87=3x
29=x

So the numbers are 28 and 29.