Question

Find the solutions for 5x+3y=30 and 3x+6y=12 FAST PLS i will mark !!!!!

Answer 1

Answer:

$x=\frac{48}{7}\\ y=-\frac{10}{7}$

Step-by-step explanation:

$5x+3y=30\\3x+6y=12$

Let's solve the second equation for x

$3x+6y=12\\3x=12-6y\\x=\frac{12-6y}{3}$

Now let's replace this in the first equation.

$5x+3y=30\\5(\frac{12-6y}{3})+3y=30$

Distribute the 5

$\frac{60-30y}{3}+3y=30$

Individually solve the fractions.

$\frac{60}{3}-\frac{30}{3}y+3y=30$

Solve;

$20-10y+3y=30\\20-7y=30$

Subtract 20 from both sides.

$20-20-7y=30-20\\-7y=10$

Divide both sides by -7

$\frac{-7y}{-7}=\frac{10}{-7}$

$y=-\frac{10}{7}$

Now replace the value of y in any of the two equations to find x.

$3x+6y=12\\3x+6(-\frac{10}{7})=12$

$3x-\frac{60}{7}=12$

add $\frac{60}{7}$ on both sides.

$3x-\frac{60}{7}+\frac{60}{7} =12+\frac{60}{7}$

$3x=\frac{(12)(7)+60}{7}$

$3x=\frac{84+60}{7}\\ 3x=\frac{144}{7}$

Multiply by the reciprocal fraction of 3. I suppose you know that 3 is 3/1 hence the inverted fraction would be 1/3

$(\frac{1}{3})(3x)=(\frac{144}{7})(\frac{1}{3})$

$x=\frac{48}{7}$