The systems of two equations: answer key to the exercises in 2de.

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Answers to the math exercises in 2nd grade on systems of equations with two unknowns. Solve with the method by linear combination or substitution.

Exercise 1:

We obtain the following system:

It can take the following different possibilities:

– 3 sets of guppies and set of neon lights: 30 + 5×4=50 cm

– 2 sets of guppies and 3 sets of neon lights; 20 + 3x5x4=80 cm

– 1 set of guppy and 4 sets of neon lights; 10+4x5x4= 90 cm

– o lot of guppy and 6 lots of neons; 6x5x4=120 cm

Conclusion: she can either put 3 sets of guppies and one set of neon lights or

2 sets of guppies and 3 sets of neon lights.

Exercise 2:

Let x be the first number and y the second number, we obtain the following system:

$\,\{\,3x+y=2\\x+2y=9\,.$

$\,\{\,3x+y=2\\x+2y=9(\times \,3)\,.$

$\,\{\,3x+y=2\\3x+6y=27\,.$

$\,\{\,3x+y=2\\(3x+y)-(3x+6y)=2-27\,.$

$\,\{\,3x+y=2\\3x+y-3x-6y=-25\,.$

$\,\{\,3x+y=2\\y-6y=-25\,.$

$\,\{\,3x+y=2\\-5y=-25\,.$

$\,\{\,3x+y=2\\y=\frac{-25}{-5}\,.$

$\,\{\,3x+y=2\\y=5\,.$

$\,\{\,3x+5=2\\y=5\,.$

$\,\{\,3x=-3\\y=5\,.$

$\,\{\,x=\frac{-3}{3}\\y=5\,.$

$\,\{\,x=-1\\y=5\,.$

Conclusion : the two chosen numbers are – 1 and 5 .

Exercise 3:

Let f be this affine function,

we have f(3)=17 and f(7)=33.

$a=\frac{f(7)-f(3)}{7-3}=\frac{33-17}{7-3}=\frac{16}{4}=4$

$b=f(3)-a\times \,3=17-4\times \,3=17-12=5$

so the affine function is

f(x)=4x+5

Let’s solve the following equations:

f(x)=37

4x+5=37

4x=37-5

4x=32

x=32:4

x=8

f(x)=9

4x+5=9

4x=9-5

4x=4

x=4:4

x=1

f(x)=17

4x+5=17

4x=17-5

4x=12

x=12:4

x=3

f(x)=29

4x+5=29

4x=29-5

4x=24

x=24:4

x=6

Conclusion: the bank card code is 8136.

Exercise 4:

Find the price of each item.

Let x be the price of a pencil and y the price of an eraser.

We obtain the following system:

$\,\{\,5x+2y=10,9\\8x+3y=17,2\,.$

$\,\{\,5x+2y=10,9\,(\times \,3)\\8x+3y=17,2\,(\times \,2).$

$\,\{\,15x+6y=32,7\,\\16x+6y=34,4\,.$

$\,\{\,15x+6y=32,7\,\\(15x+6y)-(16x+6y)=32,7-34,4\,.$

$\,\{\,15x+6y=32,7\,\\15x+6y-16x-6y=-1,7\,.$

$\,\{\,15x+6y=32,7\,\\15x-16x=-1,7\,.$

$\,\{\,15x+6y=32,7\,\\-x=-1,7\,.$

$\,\{\,15\times \,1,7+6y=32,7\,\\x=1,7\,.$

$\,\{\,25,5+6y=32,7\,\\x=1,7\,.$

$\,\{\,6y=32,7-25,5\,\\x=1,7\,.$

$\,\{\,6y=7,2\,\\x=1,7\,.$

$\,\{\,y=\frac{7,2}{6}\,\\x=1,7\,.$

$\,\{\,y=1,2\,\\x=1,7\,.$

Conclusion:

the price of a pencil is 1,4 euro and the price of an eraser is 1,2 euro .

Exercise 5:

Let x be the price of a croissant and y the price of a loaf of bread.

We obtain the following system of equations:

$\,\{\,3x+y=4,05\\4x+3y=6,9\,.$

$\,\{\,y=4,05-3x\\4x+3(4,05-3x)=6,9\,.$

$\,\{\,y=4,05-3x\\4x+12,15-9x=6,9\,.$

$\,\{\,y=4,05-3x\\12,15-5x=6,9\,.$

$\,\{\,y=4,05-3x\\-5x=6,9-\,12,15.$

$\,\{\,y=4,05-3x\\-5x=-5,25.$

$\,\{\,y=4,05-3x\\x=\frac{-5,25}{-5}.$

$\,\{\,y=4,05-3x\\x=1,05.$

$\,\{\,y=4,05-3\times \,1,05\\x=1,05.$

$\,\{\,y=0,9\\x=1,05.$

Conclusion: the price of a croissant is 1,05 euro and the price of a bread with milk is 0,90 euro.

Exercise 6:

A truck carries 20 boxes of different masses:

some weigh 28 kg, others 16 kg.

Knowing that the total mass of these boxes is 416 kg .

How many cases of each category are there?

Let x be the number of 28 kg cases and y the number of 16 kg cases.

$\,\{\,x+y=20\\28x+16y=416\,.$

$\,\{\,y=20-x\\28x+16(20-x)=416\,.$

$\,\{\,y=20-x\\28x+16\times \,20-16x=416\,.$

$\,\{\,y=20-x\\12x+320=416\,.$

$\,\{\,y=20-x\\12x=416-320\,.$

$\,\{\,y=20-x\\12x=96\,.$

$\,\{\,y=20-x=20-8=12\\x=\frac{96}{12}=8\,.$

Conclusion:

This truck contains 8 boxes of 28 kg and 12 boxes of 16 kg.

Exercise 7:

How many of each kind of bird did Betty buy?
There are in fact only two equations and three unknowns.
But the fact that the unknowns are whole allows you to solve the system anyway.

Let x be the number of ducks, y the number of chickens and z the number of chicks.

Betty buys 100 birds, so: x+y+z=100
For 100€, so : $5x+y+\frac{z}{20}=100$

From the first equation: .

We replace in the second one, to obtain the relation :

$5(100-y-z)+y+\frac{z}{20}=100$

$500-5y-5z+y+\frac{z}{20}=100$

$500-4y-\frac{100}{20}z+\frac{z}{20}=100$

$-4y-\frac{99}{20}z=100-500$

$-4y=-400+\frac{99}{20}z$

$y=\frac{-400}{-4}+\frac{99}{-4\times \,20}z$

$y=100-\frac{99}{80}z$

$y=\frac{8000-99z}{80}$

And now is the time to use integers!
According to the statement, Betty buys the chicks in groups of 20.
So she buys 20, 40, 60 or 80 chicks (0 is impossible because she buys at least one chick, and 100 is impossible because otherwise she would only buy chicks, and therefore not “at least one duck and one chicken”).

And you find:
– for z=20, y=75.25 (impossible because not integer)
– for z=40, y=50.5 (impossible because not integer)
– for z=60, y=25.75 (impossible because not integer)
– for z=80, y=1

The only integer solution is therefore y=1, z=80. And x=100-y-z=19.

Betty bought 19 ducks, 1 chicken, and 4 batches of 20 (80) chicks.

Exercise 8:

A mother is 24 years older than her daughter.

In 4 years her age will be triple that of her daughter.
How old is the daughter? How old is the mother?

Let x be the age of the daughter and y the age of the mother.

$\,\{\,y=x+24\\y+4=3(x+4)\,.$

$\,\{\,y=x+24\\x+24+4=3x+12\,.$

$\,\{\,y=x+24\\x+28-3x=12\,.$

$\,\{\,y=x+24\\-2x=12-28\,.$

$\,\{\,y=x+24\\-2x=-16\,.$

$\,\{\,y=x+24\\x=\frac{-16}{-2}\,.$

$\,\{\,y=8+24\\x=8\,.$

$\,\{\,y=32\\x=8\,.$

Conclusion: the mother is 32 years old and the daughter is 8 years old.

Exercise 9:

1. Write the equations that translate the text.

Let x be the price of a CD and y the price of a comic book.

$\{\begin{matrix}\,2x+3y=3,3,\,\,\\,4x+y=4,1\,\,\end{matrix}.$

Solve the system of equations and give the price of a CD and the price of a comic book.

$\{\begin{matrix}\,2x+3y=3,3,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,2x+3(4,1-4x)=3,3,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,2x+12,3-12x=3,3,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,-10x+12,3=3,3,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,-10x=3,3-12,3,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,-10x=-9,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,x=\frac{9}{10}=0,9,\,\,\\,y=4,1-4x\,\,\end{matrix}.$

$\{\begin{matrix}\,x=0,9,\,\,\\,y=4,1-4\times ,0,9=4,1-3,6\,\,\end{matrix}.$

$\{\begin{matrix}\,x=0,9,\,\,\\,y=0,5\,\,\end{matrix}.$

The price of a CD is 0,90 € and the price 0,50 €.

A month later, the store offers a 10% discount on CDs and 15% on comics.

How much would Loïc have paid then?

For a 10% reduction, the coefficient is 0.9 and for a 15% reduction, the coefficient is 0.85.

The price of a Cd becomes 0,9×0,9=0,81€.

The price of a DVD would be 0,5×0,85=0,425 €.

0,81×2+0,425×3=1,62+1,275=2,895 €

Loïc will therefore pay 2.895 euros.

Exercise 10:

System #1:

$\,\{\,3x+2y=8\\2x+5y=31\,.$

$\,\{\,3x+2y=8(\times \,2)\\2x+5y=31\,(\times \,3).$

$\,\{\,6x+4y=16\\6x+15y=93\,.$

$\,\{\,6x+4y=16\\(6x+4y)-(6x+15y)=16-93\,.$

$\,\{\,6x+4y=16\\6x+4y-6x-15y=-77\,.$

$\,\{\,6x+4y=16\\-11y=-77\,.$

$\,\{\,6x+4y=16\\y=\frac{-77}{-11}=7\,.$

$\,\{\,6x+4\times \,7=16\\y=\frac{-77}{-11}=7\,.$

$\,\{\,6x+28=16\\y=\frac{-77}{-11}=7\,.$

$\,\{\,6x=16-28\\y=\frac{-77}{-11}=7\,.$

$\,\{\,6x=-12\\y=\frac{-77}{-11}=7\,.$

$\,\{\,x=\frac{-12}{6}=-2\\y=\frac{-77}{-11}=7\,.$

Conclusion: ${\color{DarkRed}\,S=\,\{\,(-2;7)\,\,\}}$

System #2:

$\,\{\,2x+7y=-3\\-3x+2y=-8\,.$

$\,\{\,2x+7y=-3(\times \,3)\\-3x+2y=-8(\times \,2)\,.$

$\,\{\,6x+21y=-9\\-6x+4y=-16\,.$

$\,\{\,6x+21y=-9\\(6x+21y)+(-6x+4y)=-9+(-16)\,.$

$\,\{\,6x+21y=-9\\6x+21y-6x+4y=-9-16\,.$

$\,\{\,6x+21y=-9\\21y+4y=-25\,.$

$\,\{\,6x+21y=-9\\25y=-25\,.$

$\,\{\,6x+21y=-9\\y=\frac{-25}{25}=-1\,.$

$\,\{\,6x+21\times \,(-1)=-9\\y=\frac{-25}{25}=-1\,.$

$\,\{\,6x-21=-9\\y=\frac{-25}{25}=-1\,.$

$\,\{\,6x=21-9\\y=\frac{-25}{25}=-1\,.$

$\,\{\,6x=12\\y=\frac{-25}{25}=-1\,.$

$\,\{\,x=\frac{12}{6}=2\\y=\frac{-25}{25}=-1\,.$

Conclusion: ${\color{DarkRed}\,S=\,\{\,(2;-1)\,\,\}}$

Exercise 11:

Solve the following system:

$\,\{\,x+3y=10\\\,3x-y=0\,.$

$\,\{\,x=10-3y\\3(10-3y)-y=0\,.$

$\,\{\,x=10-3y\\30-9y-y=0\,.$

$\,\{\,x=10-3y\\30-10y=0\,.$

$\,\{\,x=10-3y\\-10y=-30\,.$

$\,\{\,x=10-3y\\y=3\,.$

$\,\{\,x=10-3\times \,3\\y=3\,.$

$\,\{\,x=1\\y=3\,.$

Exercise 12:

Calculate the length L and width l .

$\,\{\,L\times \,l=588\\\frac{L}{l}=\frac{4}{3}\,.$

$\,\{\,L=\frac{588}{l}\,\,(l\neq\,0)\,\\l=\frac{3}{4}L.$

$\,\{\,L=\frac{588}{\frac{3}{4}L}\,\,(l\neq\,0)\,\\l=\frac{3}{4}L.$

$\,\{\,\frac{3}{4}L^2=588\,\,(l\neq\,0)\,\\l=\frac{3}{4}L.$

$\,\{\,L^2=\frac{4\times \,588}{3}\,\,(l\neq\,0)\,\\l=\frac{3}{4}L.$

$\,\{\,L^2=784\\l=\frac{3}{4}L\,.$

$\,\{\,L=\sqrt{784}\,ou\,L=-\sqrt{784}\\l=\frac{3}{4}L\,.$

Now L is a length so it is a positive number.

$\,\{\,L=\sqrt{784}\\l=\frac{3}{4}L\,.$

$\,\{\,L=28\\l=\frac{3}{4}\times \,28\,.$

$\,\{\,L=28\\l=21\,.$

Exercise 13:

$\,\{\,2x+3y=0,5\\7x+5y=605\,.$

$\,\{\,2x+3y=0,5\,\,\times \,5\\7x+5y=605\,\,\times \,3\,.$

$\,\{\,10x+15y=2,5\\21x+15y=1815\,.$

$\,\{\,10x+15y=2,5\\(10x+15y)-21x-15y=2,5-1815\,.$

$\,\{\,10x+15y=2,5\\10x+15y-21x-15y=-1812,5\,.$

$\,\{\,10x+15y=2,5\\-11x=-1812,5\,.$

$\,\{\,10x+15y=2,5\\x=\frac{-1812,5}{-11}\,.$

$\,\{\,10x+15y=2,5\\x=\frac{1812,5}{11}\,.$

$\,\{\,10\times \,\frac{1812,5}{11}+15y=2,5\\x=\frac{1812,5}{11}\,.$

$\,\{\,\frac{18125}{11}+15y=2,5\\x=\frac{1812,5}{11}\,.$

$\,\{\,15y=2,5-\frac{18125}{11}\\x=\frac{1812,5}{11}\,.$

$\,\{\,15y=\frac{2,5\times \,11}{11}-\frac{18125}{11}\\x=\frac{1812,5}{11}\,.$

$\,\{\,15y=\frac{27,5}{11}-\frac{18125}{11}\\x=\frac{1812,5}{11}\,.$

$\,\{\,15y=-\frac{18097,5}{11}\\x=\frac{1812,5}{11}\,.$

$\,\{\,y=-\frac{18097,5}{11\times \,15}\\x=\frac{1812,5}{11}\,.$

$\,\{\,y=-\frac{18097,5}{165}\\x=\frac{1812,5}{11}\,.$

Exercise 14:

What is the price per kilogram of varnish and per liter of wax?

Let x : the price of one kilogram of varnish in euro and y : the price of one liter of wax in euro.

We obtain the following system:

$\,\{\,6x+4y=95\\3x+3y=55,5\,.$

$\,\{\,6x+4y=95\\3x+3y=55,5\,(\times \,2).$

$\,\{\,6x+4y=95\\6x+6y=111\,.$

$\,\{\,6x+4y=95\\(6x+4y)-(6x+6y)=95-111\,.$

$\,\{\,6x+4y=95\\6x+4y-6x-6y=-16.$

$\,\{\,6x+4y=95\\-2y=-16.$

$\,\{\,6x+4y=95\\y=\frac{-16}{-2}=8.$

$\,\{\,6x+4\times \,8=95\\y=\frac{-16}{-2}=8.$

$\,\{\,6x+32=95\\y=\frac{-16}{-2}=8.$

$\,\{\,6x=95-32\\y=\frac{-16}{-2}=8.$

$\,\{\,6x=63\\y=\frac{-16}{-2}=8.$

$\,\{\,x=\frac{63}{6}=10,5\\y=\frac{-16}{-2}=8.$

Conclusion: a kilogram of varnish costs 10.50 euros and a liter of wax costs 8 euros.

Exercise 15:

Let be the grade for the test and the grade for the homework.

$\,\{\,2x+y=33\\x+2y=39\,.$

$\,\{\,y=33-2x\\x+2(33-2x)=39\,.$

$\,\{\,y=33-2x\\x+66-4x=39\,.$

$\,\{\,y=33-2x\\-3x=39-66\,.$

$\,\{\,y=33-2x\\-3x=-27\,.$

$\,\{\,y=33-2x\\x=\frac{-27}{-3}=9\,.$

$\,\{\,y=33-2\times \,9=33-18=15\\x=9.$

Conclusion : Ahmed got 9 in the test and 15 in the homework.

Exercise 16:

1) We want to calculate the flow rate, in liters per hour, of each of the fountains.

For this purpose, we assume that the previous information is translated into the system of two equations with two unknowns:

$\{{4x+3y=55\atop\,3x+4y=57}\,$

where x is the hourly flow of the first fountain and y is the hourly flow of the second fountain.

Solve the system and indicate the hourly flow rate of each of the

two fountains.

$\{\begin{matrix}\,4x+3y=55(\times ,3),\,\,\\,3x+4y=57(\times ,4)\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,12x+16y=228\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,12x+9y-(12x+16y)=165-228\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,12x+9y-12x-16y=-63\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,-7y=-63\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,y=\frac{-63}{-7}\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9y=165,\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+9\times ,9=165,\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,12x+81=165,\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,12x=165-81,\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,12x=84,\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,x=\frac{84}{12},\,\,\\,y=9\,\,\end{matrix}.$

$\{\begin{matrix}\,x=7\,\,\\,y=9\,\,\end{matrix}.$

The hourly flow rate of the first fountain is 7 liters and that of the second is 9 liters.

2) Knowing that this basin can hold 320 liters, how long will it take to fill it, if both fountains are running together for the same time?

Every hour there will be 16 liters flowing.

320:16 = 20

It will therefore take 20 hours for this pool to be filled.

Exercise 17:
Let be the number of dromedaries and the number of camels.
There are 19 caps and a camel with 2 humps:
There are 24 pairs of slippers, each camelid has 4 legs so 2 pairs of slippers:
We have to solve the following system:
$\{\begin{matrix}\,x+y=12\,\,\\,x+2y=19\,\,\end{matrix}.$
$\{\begin{matrix}\,x+y=12\,\,\\,(x+2y)-(x+y)=19-12\,\,\end{matrix}.$
$\{\begin{matrix}\,x+y=12\,\,\\,x+2y-x-y=7\,\,\end{matrix}.$
$\{\begin{matrix}\,x+y=12\,\,\\,y=7\,\,\end{matrix}.$
$\{\begin{matrix}\,x+7=12\,\,\\,y=7\,\,\end{matrix}.$
$\{\begin{matrix}\,x=12-7\,\,\\,y=7\,\,\end{matrix}.$
$\{\begin{matrix}\,x=5\,\,\\,y=7\,\,\end{matrix}.$
Conclusion: there are 5 dromedaries and 7 camels.
Exercise 18:

$\{\begin{matrix}\,5x+2y=10,9\,(\times ,3),\,\,\\,8x+3y=17,2\,(\times ,2),\,\,\end{matrix}.$

$\{\begin{matrix}\,15x+6y=32,7,\,\,\\,16x+6y=34,4\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,15x+6y=32,7,\,\,\\,15x+6y-(16x+6y)=32,7-34,4\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,15x+6y=32,7,\,\,\\,15x+6y-16x-6y=-1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,15x+6y=32,7,\,\,\\,-x=-1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,15x+6y=32,7,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,15\times ,1,7+6y=32,7,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,25,5+6y=32,7,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,6y=32,7-25,5,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,6y=7,2,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

$\{\begin{matrix}\,y=\frac{7,2}{6}=1,2,\,\,\\,x=1,7\,,\,\,\end{matrix}.$

2)Let x be the number of pencils and y the number of erasers.

We obtain the previous system after translation of the statement.

Conclusion: the price of a pencil is €1.20 and the price of an eraser is €1.7.

Exercise 19:

Exercise 20:

Exercise 21:

Let x: the number of cars
and y: the number of motorcycles.

Mathematical translation :
There are a total of 65 vehicles so x+y = 65
There are 180 wheels so 4x+2y = 180
because a car has 4 wheels and a motorcycle has 2.
The problem amounts to solving the following system:

Conclusion: there are 25 cars and 40 motorcycles.
Exercise 22:
$\,\{{2x+3y=5,5\atop\,3x+y=4,05}$ .

1. Is the couple (x=2;y=0.5) a solution of this system ?

$\,\{{2\times 2+3\times 0,5=5,5\atop\,3\times 2+0,5=6,5}$

so this couple is not a solution of the system.

2. Solve this system of equations.

$\,\{{2x+3y=5,5\atop\,3x+y=4,05}$

$\,\{{2x+3y=5,5\atop\,y=4,05-3x}$

$\,\{{2x+3(4,05-3x)=5,5\atop\,y=4,05-3x}$

$\,\{{2x+12,15-9x=5,5\atop\,y=4,05-3x}$

$\,\{{-7x=-12,15+5,5\atop\,y=4,05-3x}$

$\,\{{-7x=-4,65\atop\,y=4,05-3x}$

$\,\{{x=\frac{-4,65}{-7}\atop\,y=4,05-3x}$

$\,\{{x=\frac{4,65}{7}\atop\,y=4,05-3\times \frac{4,65}{7}=\frac{14,4}{7}}$

3. At the bakery, Anatole buys 2 croissants and 3 pains au chocolat : he pays 5,50.

Beatrice buys 3 croissants and 1 pain au chocolat and pays 4,05 .

What is the price of a croissant? What is the price of a pain au chocolat?

By noting x: the price of a croissant and y: the price of a pain au chocolat, we arrive at the previous system.

The answer key to the math exercises on systems of two equations with two unknowns in grade 2.

After having consulted the answer key of this exercise systems of two equations in 2de, you can return to the exercises in seconde.

Second grade exercises.

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