My teacher bought 20books from sale , she spent total $110 , some books are of $3 & some are of $8 , howmany books are of $8 if the no. of books price $3 is x

Please solve it for me this from topic "Constructing Equations in Algebra To Solve Problems

My teacher bought 20books from sale , she spent total $110 , some books are of $3 & some are of $8 , howmany books are of $8 if the no. of books price $3 is x

To answer this question I will use a system of equations with two variables. If the amount of $3 books = x as you stated, then i can let the amount of $8 books = y The total amoubt of books as stated in the problem is 20 books. Our first equation would be: x+y=20. Our second equation would be a cost equation involving the total cost of the books which is stated in the problem as $110. this cost is made up ox amount of $3 books plus y amount of $8 books. the second equation would be 3x +8y = 110. I have now created a sustem of two equations. to solve this system I will use the substitution method. First I will solve one of the equations for one of the variable. The easy equation to solve is x+y =20, To solve for x I would subtract y from both sides.

x+y=20

-y = -y

x =20-y This leaves me with an answer of x = the quantity (20-y). Next I can plug that answer into my cost equation for x. The cost equation would now look like this 3(20-y) +8y=$110. We then distribute the 3 throughout the parenthesis to get: 60-3y + 8y= 110. I then combine like terms and come up with

5y +60=110. Now I will subtract 60 from both sides that leaves me with 5y = 110-60 which equals 5y=50 I will now divide both side by 5 to solve for y. 5y/5 = 50/5 that gives me y =10. Now I will plug 10 in for y into my simple equation x+10 = 20. Subtracting 10 from both sides I get x=10. To check for correctness I plug my answers into the cost equation. 3(10)+8(10) =110 = 30+80=110, 110=110.

The answer is your teacher bought 10 books at $3 each and 10 books at $8 each.

Remember to always go back and read the question again to make sure you answer it correctly. Also, most of the time x and y will not compute to the same answer,but once in a while ( as in thiis case) they do.