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# 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

### 2 Answers by Expert Tutors

Bill L. | Patient and works well with students.Patient and works well with students.
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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

Because she bought a total of 20 books:

20-x = number of \$8 books purchased

then because she spent \$110:

3x + 8(20-x) = 110

3x + 160-8x = 110

160-5x = 110

-5x = -50

x = 10  (number of \$3 books)

Number of \$8 books:

20-x = 20-10 = 10

Dan M. | Tutoring that Builds Confidence and UnderstandingTutoring that Builds Confidence and Unde...
4.9 4.9 (64 lesson ratings) (64)
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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.

Best Regards,

Dan M.

Wyzant Certified Tutor

Willoughby, OH 44094