Given: #3x-2y=0# equation 1)

#5x+10y=4# equation 2)

We will need to multiply one or both of the equations to obtain like terms in the two so we can then add or subtract one equation from the other to eliminate the like terms.

Lets try that one step at a time. We will multiply 1) by #5#:

#3x-2y=0=>5*(3x-2y=0)=>15x-10y=0#

Then **add** the new 1) to 2) to eliminate like terms:

#15xcancel(-10y)=0#

#(5xcancel(+10y)=4)/#

#20x color(white)( ........) = 4#

#5x=1 => x=1/5#

#x=1/5# and using this information we could use either 1) or 2) to solve for #y.#

But we will instead solve for #y# by elimination as well.

We will multiply 1) by #5# again:

#3x-2y=0=>5(3x-2y=0)=>15x-10y=0#

We will also multiply 2) by #3#:

#5x+10y=4=>3(5x+10y=4)=>15x+30y=12#

Now we need to **subtract** 2) from 1):

#cancel(15x)color(white)( ........) -10y=color(white)( .....) 0#

#(cancel(-15x)cancel(+)-30y=-12)/#

#color(white)( ................) -40y=-12#

#-10y=-3=>y=3/10#

To check the answers we will substitute them back into either #given# equation.

#3x-2y=0#

#3(1/5)-2(3/10)=0#

#3/5-6/10=0#

#3/5-3/5=0#