Answer:
[tex]x=\frac{48}{7}\\ y=-\frac{10}{7}[/tex]
Step-by-step explanation:
[tex]5x+3y=30\\3x+6y=12[/tex]
Let's solve the second equation for x
[tex]3x+6y=12\\3x=12-6y\\x=\frac{12-6y}{3}[/tex]
Now let's replace this in the first equation.
[tex]5x+3y=30\\5(\frac{12-6y}{3})+3y=30[/tex]
Distribute the 5
[tex]\frac{60-30y}{3}+3y=30[/tex]
Individually solve the fractions.
[tex]\frac{60}{3}-\frac{30}{3}y+3y=30[/tex]
Solve;
[tex]20-10y+3y=30\\20-7y=30[/tex]
Subtract 20 from both sides.
[tex]20-20-7y=30-20\\-7y=10[/tex]
Divide both sides by -7
[tex]\frac{-7y}{-7}=\frac{10}{-7}[/tex]
[tex]y=-\frac{10}{7}[/tex]
Now replace the value of y in any of the two equations to find x.
[tex]3x+6y=12\\3x+6(-\frac{10}{7})=12[/tex]
[tex]3x-\frac{60}{7}=12\\[/tex]
add [tex]\frac{60}{7}[/tex] on both sides.
[tex]3x-\frac{60}{7}+\frac{60}{7} =12+\frac{60}{7}[/tex]
[tex]3x=\frac{(12)(7)+60}{7}[/tex]
[tex]3x=\frac{84+60}{7}\\ 3x=\frac{144}{7}[/tex]
Multiply by the reciprocal fraction of 3. I suppose you know that 3 is 3/1 hence the inverted fraction would be 1/3
[tex](\frac{1}{3})(3x)=(\frac{144}{7})(\frac{1}{3})[/tex]
[tex]x=\frac{48}{7}[/tex]
