How do you solve the following system: 3y+x=3,5xy=14?

2 Answers
May 15, 2018

x=2(1316),y=116

Explanation:

x3y=3, Eqn (1)

5xy=14, Eqn (2)#

5 * Eqn(1) + Eqn (2) is

5x15y5xy=15+14

16y=1

y=116

Substituting value of y in Eqn (1),

x316=3

x=3+316=2(1316)

May 15, 2018

x=4516 , or 2.8125
y = 116

Explanation:

Here's our system:

3y+x=3
5xy=14

Solving By Substitution

First, let's solve for a variable. I'll choose x, since it appears first. We'll solve for x by using the first equation:

3y+x=3

Add 3y to both sides in order to negate -3y. You should now have:

x=3y3

Now, substitute this value in the second equation:

5(3y3)y=14

Distribute -5 to all terms in the parentheses. Remember negative and positive multiplication rules. (Two negatives make a positive!)

15y+15y=14

Now, combine like terms.

16y+15=14

Now, subtract 15 from both sides in order to solve for y.

16y=1

Now, divide by 16 to isolate for y.

116 = y

Because two negatives make a positive, y becomes 116.

Now, plug y in the simplified equation used to solve for x earlier:

x=3y3

Substitute y for y's value.

x=3(116)3

Multiply 3 by 1/16 to get 3/16.

x=(316)3

x=4516 , or 2.8125