Question 537186
To solve a system of two linear equations by substitution, you solve for one of the variables in one equation and substitute the expression found in the other.
Let's see how it works for your problems.
1) {{{x+3y=8}}} 
{{{y= 2x-9}}}
Half of the work is already done for you because {{{y= 2x-9}}} is already "solved" for {{{y}}}, so you just substitute {{{2x-9}}} for {{{y}}} in the other equation.
{{{x+3(2x-9)=8}}}
Multiply (apply distributive property)
{{{x+6x-27=8}}}
Collect like terms (distributive property backwards, adding apples to apples and x's to x's)
{{{7x-27=8}}}
Add 27 to both sides
{{{7x=8+27}}}-->{{{7x=35}}}
Divide both sides by 7
{{{x=5}}}
Then go back to find y
{{{y=2x-9=2*5-9=1}}}
So the answer is x=5 with y=1, usually expressed as the pair (5, 1).
2) {{{5x+2y=0}}}
   {{{x-3y=0}}}
You need to solve for x or for y one of the two equations I would go with x in the second one:
{{{x-3y=0}}}-->{{{x=3y}}}
Then substitute in the first equation
{{{5(3y)+2y=0}}}-->{{{15y+2y=0}}}-->{{{15y=0}}}-->{{{y=0}}}
and then {{{x=3y=3*0=0}}}
The solution is (0, 0).