Algebra-Simultaneous-Equations

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Algebra Linear Equations
Steps in solving
•
Forming an Equation from a statement
•
Solving for single variable
•
Solving for more than 1 variable
•
Checking using substitution
Problem 1
Three years ago, a father was 24 years older than his son. At present the father is 5 times
as old as the son. How old will the son be three years from now?
Solution:
Let the present age of son be x.
Age of father = 5x (Since it is given that the father is 5 times as old as his son)
Three years ago
Age of son = x – 3
Age of father = 5x – 3
It is given that 3 years ago father was 24 years older than his son
So,
x – 3 + 24 = 5x – 3
24 = 4x
x = 24/4 = 6 years
Present age of son = 6 years
Present age of father = 6*5 = 30 years
Age of son after 3 years = 9 years
Age of father after 3 years = 33 years
Problem 2:
The symphony sells two kinds of tickets: orchestra, for $40, and upper tiers, for $25. On a
certain night, the symphony sells 90 tickets and gets $2625 in revenue from the sales. How
many orchestra tickets did they sell?
Solution:
Let’s assume the number of orchestra tickets sold are x and number of upper tier tickets
sold are y
We know that total tickets sold = 90
i.e. x + y = 90
=> y = 90 – x
Revenue from 1 orchestra ticket = $40
Revenue from x orchestra ticket = 40x
Revenue from 1 upper tier ticket = $25
Revenue from y upper tier ticket = 25y
Total revenue = $2625
40x + 25y = 2625
40x + 25 (90 – x) = 2625
40x + 2250 – 25x = 2625
15x = 2625 – 2250 = 375
x = 375/15 = 25
total number of orchestra tickets sold = 25
Problem 3
x, y and z are internal angles of a triangle. x is twice as large as y and z is 3/4 th of the angle
x. Find the angles.
Solution:
Given x is twice as large as y
i.e. x = 2y => x/2 = y
z is 3/4th of the angle x
z=¾*x
We know that the sum of internal angles of a triangle = 1800
x + y + z = 180
Replacing y and z by their values in term of x
x + x/2 + ¾ * x = 180
Multiplying by 4
4x + 2x + 3x = 720
9x = 720
x = 800
y = x/2 = 80/2 = 400
z = 3x/4 = 3*80/4 = 3*20 = 600
Simultaneous Equations:
Simultaneous Equations are set of equations with one or more variables. They are solved
using a set of operations. There are several ways to solve simultaneous equation, most
basic of them uses substitution.
Note: The equations may or may not have consistent solution. And Equations may have
infinite solutions.
Let’s check out with an example:
•
Equation 1: x - y = -10
Equation 2: 2x + 3y = 15
Steps to solve:
Method 1:
Equation 1, Rearranged: y = x + 10 (Finding Value of y in terms of x)
Value of y substituted in equation 1: 2x + 3(x + 10) = 15
2x + 3x + 30 = 15
5x + 30 = 15
5x + 30 - 30 = 15 - 30
5x = -15
x = -3
Substituting x in 1 to solve for y: 2(-3) + 3(y) = 15
-6 + 3y = 15
3y = 21
y=7
Solution: x = -3, y = 7
Method 2: Multiplying both the equations with a constant and eliminating a
variable by adding or subtracting
For the same two equations
Multiply equation 1 by 3
We get, 3x – 3y = -30 … (a)
Equation 2
2x + 3y = 15 … (b)
Adding (a) and (b)
2x + 3x – 3y + 3y = -30 + 15
5x = -15
x = -3
substituting value of x in equation 1
x – y = -10
-3 – y = -10
y = -3 + 10 = 7
Method 1 is usually considered easier to understand but Method 2 is quicker
once we get complex problems
Problem 1:
2x+3y=1
7x+6y=8
Using method 2
Multiply equation 1 by 7
14x + 21y = 7 …(a)
Multiply equation 2 by 2
14x + 12y = 16 … (b)
Subtracting (b) from (a) i.e. following (a) – (b)
14x – 14x + 21y – 12y = 7 – 16
9y = -9
y = -1
substituting y = -1 in equation 1
2x + 3(-1) = 1
2x – 3 = 1
2x = 4
x=2
Problem 2
Machines A and B always operate independently at their respective constant rates. When
working alone, machine A can fill a production lot in 5 hours, and machine B can fill the
same lot in x hours. When the two machines operate simultaneously to fill the production
lot, it takes them 2 hours to complete the job. What is the value of x?
Solution:
Let the total amount of work be W
Machine A can finish the work in 5 hours
Rate (Per hour) at which machine A works = W/5
Similarly, Rate at which Machine B works = W/x
Rate at which both simultaneously work = W/5 + W/x
They finish the work in 2 hours
Total work done = rate * time
W = (W/5 + W/x) * 2
W = W (1/5 + 1/x) * 2
1 = (1/5 + 1/x) * 2
½ = 1/5 + 1/x
1/x = ½ - 1/5 = 3/10
x = 10/3 hours
Problem 4
One woman and one man can build a wall together in two hours, but the woman would
need the help of two girls to complete the same job in the same amount of time. If one
man and one girl worked together, it would take them four hours to build the wall. If rates
for men, women and girls remain constant, how many hours would it take one woman,
one man, and one girl, working together, to build the wall?
Solution:
Let the rate at which a man, a woman and a girl builds the wall be m, w and g respectively
Let the total work be 1 unit
We know that one man and one woman can finish the work in 2 hours
Total rate of man and woman = m + w
Rate * time = work
(m + w) * 2 = 1
2m + 2w = 1 … (a)
A woman and two girls finish the same job in 2 hours
i.e.
2*(w + 2g) = 1
2w + 4g = 1 … (b)
Also 1 man and 1 girl finish the same job in 4 hours
i.e.
4*(m + g) = 1
4m + 4g = 1 … (c)
Solving for m, w and g
(a) - (b)
2m – 4g = 0
m = 2g
replacing m = 2g in (c)
4(2g) + 4g = 1
10g = 1
g = 1/10
Substituting g in (c)
4m + 4/10 = 1
4m = 6/10 = 3/5
m = 3/20
substituting g in (b)
2w + 4/10 = 1
2w = 6/10
w = 3/10
combined rate of 1 man, 1 woman and 1 girl
=m+w+g
= 3/20 + 3/10 + 1/10 = 11/20
Time taken = work / rate = 1 / (11/20) = 20/11 hours
Problem 3
Working alone at their respective constant rates, A can complete a task in ‘a’ days and B in
‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts
they finish the task in exactly 10 days. If B starts, they take half a day more. How long does
it take to complete the task if they both work together?
Solution
Let the total work be 1 unit
Rate (per day) at which A does work = 1/a
Rate (per day) at which B does work = 1/b
If A starts they finish the work in 10 days
Since each work 2 days the amount of work done will be rate*time
2/a + 2/b + 2/a + 2/b + 2/a = 1 (Total work)
6/a + 4/b = 1 … (p)
If B starts they finish the work in 10.5 days
Since each work 2 days the amount of work done will be rate*time
2/b + 2/a + 2/b + 2/a + 2/b + 0.5/a = 1
4.5/a + 6/b = 1
Multiplying by 2
9/a + 12/b = 2 … (q)
Multiplying (p) by 3 and then (p) – (q)
9/a = 1
a=9
substituting a in (q)
9/9 + 12/b = 2
12/b = 1
b = 12
Combined rate of work = 1/a + 1/b = 1/9 + 1/12
Let total time taken be t
Since rate * time = total work
(1/9 + 1/12) * t = 1
7/36 * t = 1
t = 36/7 days approx. 5.1 days
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