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Solving Equations help.

[From: ] [author: ] [Date: 11-08-23] [Hit: ]
Solve by the substitution method. 15x - 20y = -1 15y = 2 + 5x3. Solve by the elimination method. 8x - y = 68x + 3y = 214. Solve by the substitution method. 15x - 10y = -5 15y = 4 + 5x5.......
1. The product of two consecutive positive integers is 11 more than their sum. Find the integers.

2. Solve.

v^2 = 18 - 3v

3. Solve by the substitution method.

15x - 20y = -1
15y = 2 + 5x

3. Solve by the elimination method.

8x - y = 68
x + 3y = 21

4. Solve by the substitution method.

15x - 10y = -5
15y = 4 + 5x

5. Solve the system of equations by the substitution.

0.06x - 0.01y = 0.6
0.22x + 0.13y = 12.2

6. A motel clerk counts his $1 and $10 bills at the end of a day. He finds that he has a total of 59 bills having a combined monetary value of $239. Find the number of bills of each denomination that he has.

the clerk has ____ ones and ___ tens.

-
1)

the first integer is = x
the second integer is ( x + 1 )

the product as x (x + 1)
their sum x + (x + 1) -----> 2x + 1

consecutive positive integers is 11 more than their sum
x (x + 1) = 11 + 2x + 1
x^2 + x = 12 + 2x
x^2 + x - 12 - 2x = 0
x^2 - x - 12 = 0
(x - 4)(x + 3) = 0 =====> x = -3 & 4

==========

2)

v^2 = 18 - 3v

v^2 + 3v - 18 = 0

(v + 6)(v - 3) = 0 =====> v = -6 & 3

=========

3)

15x - 20y = -1 ====> 15x = 20y - 1 =====> x = (20/15) y - (1/15) ===> x = (4/3) y - (1/15)

substitute it into :
15y = 2 + 5 * ( (4/3) y - (1/15) )
15y = 2 + (20/3) y - (1/3)
15y - (20/3) y = (5/3)
(25/3) y = (5/3)
25 y = 5
y = 1/5

when y = 1/5
x = (4/3) (1/5) - (1/15)
x = (4/15) - (1/15)
x = (3/15)
x = 1/5

============

3)

8x - y = 68
x + 3y = 21 ===> * -8


8x - y = 68
+
-8x - 24y = -168
-----------------------
0 - 25y = -100 ======> y = 4

when y = 4
8x - 4 = 68
8x = 68 + 4
8x = 72
x = 9

===========

4)

15x - 10y = -5 ====> 15x = 10y - 5 ====> x = (10/15) y - (5/15) ====> x = (2/3) y - (1/3)
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