cadaVR anatomy by Mozaik Education

Simultaneous equations (systems of equations)

ALGEBRAIC METHODS FOR SOLVING A LINEAR SYSTEM OF EQUATIONS WITH TWO VARIABLES:

Substitution;
Equating coefficients.

Example 1

The organizers of a prom have to seat \latex{ 470 } guests at tables for \latex{ 4 } and \latex{ 6 }. How many tables are needed from each type if there is enough space for \latex{ 100 } tables overall?

Solution

Let the number of tables for \latex{ 4 } be \latex{ x }, that of tables for \latex{ 6 } be \latex{ y }. By the criterion concerning the total number of tables, \latex{x + y = 100}. By the criterion concerning the number of guests, \latex{4x + 6y = 470.}

We need to solve the following system of equations:

\latex{\begin{rcases}x+y=100 \\ 4x+6y=470 \end{rcases}.}

Subtract the \latex{ 4 }-times of the first equation from the second:

\latex{\begin{rcases}4x+6y=470 \\ 4x+4y=400\end{rcases}} ,
\latex{2y=70,}
\latex{y=35}

By substituting into the first equation, \latex{x = 65.}

It can be checked by simple calculation that \latex{ 35 } tables for \latex{ 6 } and \latex{ 65 } tables of \latex{ 4 } are needed to seat the guests.

Example 2

A two-digit number is smaller by \latex{ 36 } than the number obtained by reversing the order of its digits. If divided by the sum of its digits, the quotient will be the first digit of the original number and the remainder will be \latex{ 7 }. What is the two-digit number in question?

Solution

Let the first digit be \latex{ x } and the second one be \latex{ y }. The value of the \latex{ 2 }-digit number \latex{\overline{xy}} is \latex{10x+y.} After reversion \latex{10y+x=10x+y+36.} From the criterion concerning the division: \latex{10x+y=x\times (x+y)+7.}

We have to solve the following system of equations:

\latex{\begin{rcases}10y+x=10x+y+36 \\ 10x+y=x\times (x+y)+7\end{rcases}.}

By rearranging the first equation: \latex{9y = 9x+36}, which yields: \latex{y = x + 4.}

By substituting into the second one gives \latex{10x+x+4=x\times(2x+4)+7.}

After expanding the parentheses and reordering, we have

\latex{11x+4=2x^{2}+4x+7,}
\latex{0=2x^{2}-7x+3.}

Solving by the quadratic formula,

\latex{x_{1}=3} and \latex{x_{2}=\frac{1}{2}.}

As we are looking for a digit, the only possibility is \latex{ 3 }, which gives \latex{y = 7} after substitution.

The number in question is \latex{ 37 }. Indeed, it is smaller by \latex{ 36 } then \latex{ 73 } and if divided by \latex{ 10 }, the quotient is \latex{ 3 }, while the remainder is \latex{ 7 }.

Example 3

The sum of two numbers is \latex{ 1 }. If the double of the reciprocal of one number is added to \latex{ 3 } times the reciprocal of the other number, the result will be \latex{ 12 }. What are the two numbers?

Solution

Let the two numbers be \latex{ x } and \latex{ y }.

The two criteria give us the following system of equations:

\latex{\begin{rcases}x+y=1 \\ \frac{2}{x}+\frac{3}{y}=12\end{rcases}.}

Express \latex{ y } from the first one \latex{(y =1– x)}, and substitute it into the second one:

\latex{\frac{2}{x}+\frac{3}{1-x}=12.}

After multiplication and ordering,

\latex{2\times (1-x)+3=12x\times (1-x),}
\latex{2-2x+3x=12x-12x^{2},}
\latex{12x^{2}-11x+2=0.}

Using the quadratic formula gives

\latex{x_{1}=\frac{2}{3}} and \latex{x_{2}=\frac{1}{4}}.

The corresponding values of \latex{ y } are:

\latex{x_{1}=\frac{1}{3}} and \latex{y_{2}=\frac{3}{4}}.

It can be checked by easy calculation that both pairs of numbers are solutions.

Example 4

Solve the following system of equations on the set of pairs of real numbers:

\latex{\begin{rcases}\sqrt{x-y+5}=3 \\ \sqrt{x+y-5}=-2x+11\end{rcases}.}

Solution

The system of equations can be solved only if \latex{x-y+5\geq 0} and \latex{x+y-5\geq 0} holds.
Take the squares of both sides of both equations:

\latex{\begin{rcases}x-y+5=9 \\ x+y-5=4x^{2}-44x+121\end{rcases}.}

By adding the two equations, variable \latex{ y } will disappear:

\latex{2x=4x^{2}-44x+130.}

After division by \latex{ 2 } and reordering, we have

\latex{0=2x^{2}-23x+65.}

Its solution, using the quadratic formula, is

\latex{x_{1}=6.5} and \latex{x_{2}=5}.

Substitution into equation (1) yields

\latex{y_{1}=2.5} and \latex{x_{2}=1}.

Both pairs satisfy the criteria concerning the domain, but substitution into the original system of equations show that only the second pair is a solution.

Therefore the solution of the system of equations is the pair (\latex{ 5 }; \latex{ 1 }).

Example 5

Solve and examine the following system of equations, if \latex{ p } is a real parameter.

\latex{\begin{rcases}px+3y=12 \\ 12x+py=24\end{rcases}.}

Solution

Variable \latex{ y } will disappear after multiplication of the first equation by \latex{ p } and the second one by \latex{ 3 } and subtraction:

\latex{\begin{rcases}p^{2}x+3py=12p \\ 36x+3py=72\end{rcases}.}

Subtract the second equation from the first one:

\latex{x\times (p^{2}-36 )=12p-72,}
\latex{x\times (p-6)\times (p+6)=12\times (p-6).} \latex{(*)}

If \latex{p = 6}, then the equation is true for any \latex{ x }. In this case the original two equations coincide: \latex{6x + 3y = 12.} Therefore in this case every pair of numbers \latex{(x; y)} is a solution of the system of equations which satisfies the equation \latex{6x + 3y = 12.} (Represented in the Cartesian coordinate system, the set of such points form a line.) (Figure 45)

If \latex{p = –6,} then the left hand side of the equation \latex{(*)} is \latex{ 0 } while the right hand side is not. Thus in this case there is no solution.

If \latex{p\neq 6} and \latex{p\neq -6}, then from \latex{(*)} we have

\latex{x=\frac{12}{p+6},}

which gives

\latex{y=\frac{24}{p+6}}

after substitution into the first equation.

Exercises

{{exercise_number}}. The price of the half liter soft drink in the supermarket is \latex{ 139 } Euros. The \latex{ 2 } liters pack of the same drink costs \latex{ 237 } Euros. How much does one liter of the drink costs supposing that

every plastic bottle costs the same;
the two liters bottle costs twice as much as the half liter one?

Explain the results.

{{exercise_number}}. There are blue and red marbles in a box. If there would be one fewer red, then one seventh of the marbles would be red. Also, if we would replace two blue marbles by red ones, one fifth of the marbles would be red. How many red and blue marbles are there in the box?

{{exercise_number}}. Dad is rearranging the bookshelves. Until now there were \latex{ 12 } books on each shelf, but \latex{ 4 } books did not have place. After rearrangement there will be \latex{ 13 } books on each shelf except the topmost as there will be only \latex{ 8 } books on that one. How many shelves are there and how many books will be on the bookshelves after the rearrangement?

{{exercise_number}}. \latex{ 75 }% of an alloy of gold and silver is gold. This alloy worth \latex{ 190 }% more than the alloy with reverse components (\latex{ 25 }% gold, \latex{ 75 }% silver).

How many times more a unit of gold is worth compared to a unit of silver?
In percentage, how much less a unit of silver is worth compared to a unit of gold?

{{exercise_number}}. Solve the following equations on the set of pairs of real numbers.

\latex{\begin{rcases}\frac{2}{x}+\frac{1}{y}= \frac{1}{xy} \\ x^{2}+y^{2}=1\end{rcases}};

\latex{\begin{rcases}\frac{3}{x}-2y=1 \\ -2x+\frac{3}{y}=3\end{rcases}};

\latex{\begin{rcases}x^{2}+xy=210 \\ y^{2}+xy=231\end{rcases}}.

{{exercise_number}}. Solve the following equations on the set of pairs of real numbers.

\latex{\begin{rcases}2\times \sqrt{\frac{5x}{x-y} }+\sqrt{\frac{x-y}{5x} }=3 \\ (y+4x)\times (x-y+20)=0\end{rcases}};

\latex{\begin{rcases}3\times 4^{x+y}-5\times 2^{x-y}=\frac{91}{4} \\ 5\times 4^{x+y}-4\times 2^{x-y}=39\end{rcases}};

\latex{\begin{rcases}x^{4}+y^{4}=641 \\ \log x^{2}+\log y^{2}=2\end{rcases}}.

Mathematics 12.

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