Resolved Problems of Systems of Linear Equations

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Introduction
A system of equations is a group of equations (in our case it will be two) and various unknown
factors (two in our case) that appear in one or various
of these equations.
An equation with more than one unknown factor informs of the relation between them. For example, the equation x  y = 0 tells that x and y are the same number.
We cannot solve an equation with two unknown factors, because one of them is dependent on the other. For instance, if we
have the equation x  2y = 0 and we isolate x, we obtain that x = 2y.
So, the value of x is double
the value of y. But we still do not know the values of x and y.
To resolve a system of N unknown factors, we need N equations.
In fact, we need the equations to be linearly independent, but we will not take that into consideration at this level.
In this section we have exercises that require to be considered as a system
of equations of second dimension (two equations and two unknown factors). If
we do not remember how to resolve systems (equalization, reduction and substitution),
we can visit the section of Systems.
Resolved Problems
Problem 1
Two numbers add 25 and the double of one of them is 14. What numbers are they?
Show solution
We call x the first number and y the second number.
The numbers add 25:
$$ x + y = 25 $$
The double of one of the numbers is 14 (we can consider that this number is x):
$$ 2x = 14 $$
We have the system
We apply substitution:
So, the numbers are 7 and 18.
Problem 2
The double of the sum of the two numbers is 32 and the subtraction of them is 0.
What numbers are they? We apply reduction.
Show solution
We call x the first number and y the second number.
The double of the sum of two numbers is 32:
$$ 2(x + y) = 32 $$
The subtraction of them is 0:
$$ x  y = 0 $$
We have the system
We apply reduction:
So, the numbers are 8 and 8.
Problem 3
Adding two numbers gives us 12 and half of one of them is double the
other. What numbers are they?
Show solution
x = first number
y = second number
The numbers add 12:
$$ x + y = 12 $$
Half of one of them is double the other:
$$ \frac{x}{2} = 2y $$
We have the system
We apply substitution:
So, the numbers are
$$ \frac{18}{5},\ \frac{12}{5} $$
Problem 4
We have two numbers that add 0, and if we add 123 to one of them, we get the double of the other one.
What numbers are they?
Show solution
We call
x = first number
y = second number
The numebers add 0:
$$ x + y = 0 $$
If we add 123 to one of them, we get the double of the other one:
$$x + 123 = 2y $$
We have the system:
We apply substitution:
So, the numbers are 41 and 41.
Problem 5
Find the two digit number so that:
the second digit is double the first and adding both digits
gives us 12.
Show solution
The number is xy,
where x is the first digit and y is the second one.
The second digit is double the first:
$$ y = 2x $$
Adding both digits gives 12:
$$ x + y = 12 $$
We have the system:
We resolve with substitution:
So, the number is 48.
Problem 6
Anne is three times the age of her son Jacob.
In 15 years, Anne's age will be double her son's. How many years older is Anne than Jacob?
Show solution
We call
a =Anne's age
j = Jacob's age
Anne is three times the age of her son Jacob:
$$ a = 3j $$
In 15 years, Anne's age will be double Jacob's:
$$ ( a + 15 ) = 2( j + 15 ) $$
We have the system:
We apply substitution:
Anne is 45 years old and her son Jacob is 15 year's old,
so Anne is 30 years older than her son.
Problem 7
We have bought 3 glass marbles and 2 steel ones for 1,45$ and, yesterday, 2 glass ones and 5
steel ones for 1,7$. Determine the price of one glass
marble and a steel one.
Show solution
We call
c = number of glass marbles
a = number of steel marbles
3 glass marbles and 2 steel ones are 1.45$:
$$ 3c + 2a = 1,45 $$
2 glass marbles and 5 steel ones are 1.7$:
$$ 2c + 5a = 1,7 $$
We have the system:
We apply reduction:
One steel marble is 0.2$ and a glass one is 0.35$.
Problem 8
Work out the length of the sides of a rectangle
that its perimeter is 24 and its
large side measures three times the short side.
Show solution
The rectangle has 4 sides:
two are the same as each other (base) and another two
are the same as each other (height).
The perimeter is the sum of all the sides.
We call x the large side and y the short side.
The perimeter is 24:
$$ 2x + 2y = 24 $$
The large side measures three times the short side:
$$ x = 3y $$
We have the system:
We apply substitution:
The large sides measure 9 units and the short sides 3 (each of them).
Problem 9
Work out the number of animals in a farm knowing that:
the sum of ducks and cows is 132 and their legs add 402.
200 Kg are needed a day to feed the chickens and the roosters. There is one
rooster per 6 chickens and the average amount a chicken eats is 500g,
double what the rooster does.
It is known that the sixth part of the
rabbits escapes to the cows keep, which means there is triple
the animals in the keep.
Show solution
a. Take into account that each duck has 2 legs and
each cow has 4. The sum is 402 (two legs per duck and 4 per cow).
p = number of ducks
v = number of cows
These animals add 132:
$$ p + v = 132 $$
The sum of their legs is 402 (two legs per duck and 4 per cow):
$$ 2p + 4v = 402 $$
We have the system:
We apply reduction:
There are 63 ducks and 69 cows.
b. In accord with the quantities of food are estimated,
we won't obtain a exact number of animals, only an estimation.
y = number of roosters
x = number of chickens
There is one rooster per 6 chickens:
$$x = 6y $$
A chicken eats is 0.5Kg, double what the rooster does. The total is 200Kg:
$$0,5x + 0,25y = 200 $$
We have the system:
We apply substitution:
We can say there are 61 roosters and 366 chickens.
c. We know there are 69 cows.
c = number of rabbits
The sixth part of the rabbits are with the cows, so there are
$$ 69 + c/6 $$
animals in the cow's keep.
Counting the rabbits, the number of animals in the keep is triple:
$$ 69 + \frac{c}{6} = 69\cdot 3 $$
We resolve the linear equation:
There are 828 rabbits.
Problem 10
In a multiple choice exam, the correct questions add a point and the wrong
answers subtract half a point. In total there are 100 questions
and none can be left unanswered (all must be answered).
The student gets 8.05 out of 10 in the exam.
Work out the number of questions answered correctly and incorrectly.
Show solution
We write the score as 100 instead of 10:
$$ 8.05 \cdot 10 = 80.5 $$
We call x the number of correct answers and y the wrong answers.
Due to the fact that all questions must be answered, the following
equation must be true
$$ x + y = 100 $$
Each correct answer adds 1 and each wrong subtracts 0.5:
$$ 1\cdot x  0.5 \cdot y = 80.5 $$
Now we resolve the equation system by substitution:
We isolate the x from the first equation:
$$ x + y = 100 $$
$$ x = 100  y $$
Now we substitute x in the second equation:
$$ x  0.5 \cdot y = 80.5 $$
$$ (100y)  0.5 \cdot y = 80.5 $$
We resolve the linear equation:
$$ 100y  0.5 \cdot y = 80.5 $$
$$ 10080.5 = y +0.5 y $$
$$ 19.5 = 1.5y$$
$$ y = \frac{19.5}{1.5} = 13$$
We conclude that the amount of wrong answers is y = 13.
We can easily work out how may correct answers there are:
$$ x = 100  y = 100  13 = 87 $$
Problem 11
If we add 7 to the numerator and denominator of a fraction, we obtain this fraction
$$ \frac{2}{3} $$
If instead of adding 7 we subtract 3 from the numerator and denominator, we obtain this fraction
$$ \frac{1}{4} $$
Find said fraction.
Show solution
We call the numerator x and the denominator y.
So, the fraction is
$$ \frac{x}{y} $$
We add 7 to the numerator and denominator and we get 2/3:
$$ \frac{x+7}{y+7} = \frac{2}{3} $$
Notice that from the equality above we get:
$$ 3(x+7) = 2(y+7) $$
We operate to simplify it:
$$ 3x+21 = 2y+14 $$
$$ 3x 2y = 7 $$
Now we proceed the same way but subtracting 3:
$$ \frac{x3}{y3} = \frac{1}{4} $$
We operate to simplify it:
$$ 4(x3) = y3 $$
$$ 4x12 = y3 $$
$$ 4xy = 9 $$
Therefore the linear equation system is the following:
$$ 3x 2y = 7 $$
$$ 4xy = 9 $$
We solve it with equalization (to change the method). Thus
we isolate the x's from both equations.
From the first equation:
$$ 3x 2y = 7 $$
$$ x = \frac{7+2y}{3} $$
From the second equation:
$$ 4xy = 9 $$
$$ x = \frac{9+y}{4} $$
We equal both x's:
$$ \frac{7+2y}{3} = \frac{9+y}{4}$$
We resolve the linear equation:
$$ 4(7+2y) = 3(9+y)$$
$$ 28+8y = 27 + 3y $$
$$ 5y = 55 $$
$$ y = \frac{55}{5} = 11 $$
We calculate x from y:
$$ x = \frac{7+2y}{3} =$$
$$ x = \frac{7+22}{3} =$$
$$ x = \frac{15}{3} =$$
$$ x = 5 $$
So, the fraction we obtain is
$$ \frac{x}{y} = \frac{5}{11} $$
Problem 12
A make of soft drinks makes a lemonade (water with lemon concentrate) with a very specific
quantity of ingredients. The relation between the quantity of water and concentrate is
$$ L_L = \frac{2 L_A}{5} $$
where L_{L} represents the litres of concentrate and
L_{A} the litres of water.
If 20 lemons are needed to make one litre of lemon concentrate,
how many lemons will they need to make 1230 2L bottles of this lemonade?
Show solution
The 1230 2L bottles are equivalent to a total of
$$ 1230 \cdot 2 = 2460 L $$
litres of lemonade.
One of the equations of the system is the one provided by the exercise information.
The other equation is the following:
$$ L_L + L_A = 2460 $$
So, the total litres of lemonade is the sum of water and concentrate.
We resolve the system by substitution:
From the second equation:
$$ L_L = 2460  L_A $$
We substitute in the first equation:
$$ L_L = \frac{2 L_A}{5} $$
$$ 2460  L_A = \frac{2 L_A}{5} $$
We resolve the linear equation:
$$ 5\cdot 2460  5L_A = 2L_A $$
$$ 12300 = 5L_A + 2L_A $$
$$ 12300 = 7L_A $$
$$ L_A = \frac{12300}{7} = 1757.14L $$
Note: L indicates the litres, it has nothing
to do with the name of the unknown factor.
Now we calculate the litres of lemon concentrate:
$$ L_L = 2460  1757.14 = 702.86L $$
Since we need 20 lemons to make a litre of concentrate and we want
702.86L of concentrate, we need
$$ 20\cdot 702.86 = 14057.2 $$
The total of lemons needed is 14058.
Problem 13
High difficulty
With a 34 metre rope we can draw a rectangle (without there being any rope left over)
that has a diagonal of 13 metres.
Calculate how much the base and height of the rectangle are.
Show solution
We call b the base of the rectangle and a the height.
Note that the length of the rope is the perimeter. Therefore
$$ 2a + 2b = 34 $$
If we draw the diagonal of the rectangle, we will see two right angle
triangles, the hypotenuse being the diagonal:
We apply the Pithagoras' Theorem:
$$ a^2 + b^2 = h^2 $$
where h represents the hypotenuse (the diagonal). Then,
$$ a^2 + b^2 = 13^2 = 169 $$
The difficulty of the problem is due to the fact the last equation is not linear
(the unknown factors are to the power of two).
We isolate a from the first equation:
$$ 2a + 2b = 34 $$
$$ a = \frac{342b}{2} = 17b$$
Now we substitute a in the non lineal equation:
$$ a^2 + b^2 = 169 $$
$$ (17b)^2 + b^2 = 169 $$
We calculate the square of the subtraction (newton's binomial theorem):
$$ (17b)^2 = 17^2 +b^2 2\cdot 17b =$$
$$ = 289 +b^2 34b $$
Then, we have a quadratic equation:
$$ 289 +b^2 34b + b^2 = 169 $$
We simplify a bit:
$$ 2b^2 34b +120 =0 $$
We omit the procedure, because it does not belong to the unit of systems of equations.
The solutions are:
$$ b=12,\ b=5 $$
So a has two values:
$$ a = 17b = 175=12 $$
$$ a = 17b = 1712=5 $$
Notice that in fact there is only one solution to the problem, because if we consider b = 12
then a = 5 and if b = 12, then a = 5.
This is due to the fact is does not matter if we consider a side as the base or the height.
Therefore the sides of the rectangle are 12 and 5 meters.
Problem 14
In a charity concert they sell all the tickets and the raise 23 thousand dollars.
The price of the tickets is 50 for normal ones and 300 dollars for the V.I.P.
Work out the number of tickets of each type that were sold if the venue holds 160 people.
Show solution
We will call the number of VIP tickets v and the number of normal tickets n
(it does not matter what name we give the unknown factors).
The total number of tickets is the same as the number of people:
$$ v + n = 160 $$
They raise:
$$ 300v + 50n = 23000 $$
We resolve the system with equalization. Thus we isolate c in both equations:
From the first equation:
$$ v = 160n $$
From the second equation:
$$ v = \frac{23000  50n}{300} $$
We equal both expressions:
$$ 160n = \frac{23000  50n}{300} $$
The solution to the linear equation above is:
$$ n = 100 $$
Therefore
$$ v = 160100 = 60 $$
So, they sold 60 VIP tickets and 100 normal.
Problem 15
A child makes an observation in the ball play park:
There are green, red and yellow balls.
The number of green balls and red balls is five times the yellow ones.
The number of green ones are three times the yellow ones.
The total amount of yellow and red balls is 123.
Show solution
We will call g the number of green balls, r the number
of red balls and y the number of yellow balls.
The second point says that
$$ g+r = 5y $$
The third point says that
$$ g = 3y $$
And the fourth point says that
$$ y+r = 123 $$
We have a system of linear equations of 3 equations with 3 unknown factors.
We apply substitution:
We substitute g from the second equation in the first one:
$$ g+r = 5y $$
$$ 3y+r = 5y $$
We isolate r:
$$ r = 5a 3a $$
$$ r = 2a $$
We substitute r in the third equation:
$$ y+r = 123 $$
$$ y+2y = 123 $$
$$ 3y = 123 $$
$$ y = 41 $$
Now we use the value of y to obtain the other unknown factors:
$$ r = 2y = 82 $$
$$ g = 3y = 123 $$
Problem 16
Calculate the number of positive numbers with three digits
(bigger than 99) so that one of the digits is 0 and the other two add 7.
Show solution
Since the numbers are 3 digits long, it will look like
$$ xy0 $$
or
$$ x0y $$
where x and y represents digits.
Notice that the numbers 0xy are not really three digit long, because
0xy = xy.
In both cases, it has to be
$$ x+y = 7 $$
Hence we have these possibilities
x 
y 
x + y 
1 
6 
7 
2 
5 
7 
3 
4 
7 
4 
3 
7 
5 
2 
7 
6 
1 
7 
7 
0 
7 
Notice that x cannot be 0, because in both cases it is the first digit.
In the table we have the 7 possibilities and are valid for both cases.
So, to start with there will be 14 numbers,
but we have to remove one because the values
of the last row provide the same number.
The solution is 13 numbers.
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