Simultaneous equations (Systems of linear equations): Problems with Solutions

Solution:
We have to substitute x by 0 and y by $\frac{5}{2}$.
If we get correct equations then the point is a solution.

$5(0) +2(\frac{5}{2}) = 0 + 5 = 5 \neq 1$

The point does not satisfy the first equation, so it is not a solution.

Solution:
We substitute x by 1 and y by 3 into each of the equations.
$2(1) -3 =2 -3 = -1$
$3(1) +3 =3 +3 =6$
The point $(1, 3)$ satisfies the both equations, therefore it is a solution.

Solution:
A point [tex]\left(x , y\right)[/tex] is a solution of the system if its coordinates satisfy both equations. Check the points [tex]\left(1 , 3\right)[/tex] and [tex]\left(-1 , -1\right)[/tex]:
[tex]x=1,\ y=3:\quad 3-2\times 1=1,\qquad 6\times 1-3\times 3=-3[/tex]
[tex]x=-1,\ y=-1:\quad -1-2\times (-1)=1,\qquad 6\times (-1)-3\times (-1)=-3[/tex]
Both points satisfy both equations, so both are solutions. Each of the other pairs contains a point that does not satisfy the first equation [tex]y-2x=1[/tex]:
[tex]\left(0 , 0\right):\quad 0-2\times 0=0\ne 1[/tex]
[tex]\left(1 , 0\right):\quad 0-2\times 1=-2\ne 1[/tex]
[tex]\left(1 , -2\right):\quad -2-2\times 1=-4\ne 1[/tex]
Therefore the answer is the pair [tex]\left(1 , 3\right)[/tex] and [tex]\left(-1 , -1\right)[/tex]. In fact, the second equation is the first one multiplied by [tex]-3[/tex], so both equations describe the same line [tex]y=2x+1[/tex], and the solutions of the system are exactly the points of this line.

Solution:
Multiply the first equation by [tex]-3[/tex]:
[tex]-3(y-2x)=-3\times 1\iff 6x-3y=-3[/tex]
This is exactly the second equation, so the two equations are equivalent and their graphs coincide: both are the line [tex]y=2x+1[/tex]. Every point of this line is a solution of the system, for example [tex]\left(0 , 1\right)[/tex], [tex]\left(1 , 3\right)[/tex] and [tex]\left(-1 , -1\right)[/tex]. The solutions are all pairs of the form
[tex]\left(x , 2x+1\right),\quad x\in\mathbb{R}[/tex]
Hence the system has infinitely many solutions. A system of two linear equations in two unknowns has either exactly one solution (the lines intersect), or no solution (the lines are parallel), or infinitely many solutions (the lines coincide); it can never have exactly two solutions.

Solution:
When a system of equations has infinite solutions this does not guarantee that any point of the plane is a solution, we can verify this fact by taking the arbitrary point $(2 ,1)$ which does not satisfy any of the equations, for this reason it can not be a system solution. Now, let's verify that the previous system of equations has infinite number of solutions, note that if we multiply the first equation by 3, we get the second equation.

$(3) \ast (x +2y =1)$ $ \Longrightarrow 3x +6y =3$
This says that the two equations are equivalent, we solve one of the two equations.
$x +2y =1 \Longrightarrow x =1 -2y$
We can affirm that the points of the plane that satisfy this relation are solutions of the system (if we work with the second equation we arrive at the same result). All the points $(\mathbf{x} ,\mathbf{y}) =(1 -2\mathbf{y} ,\mathbf{y})$ that have this form are solutions to the first equation (and also of the second because they are equivalent)
By assigning any real value to $y$, we obtain the corresponding value of $x$, and so we have a solution, we can give infinite values of $y$ and thus we get the value of $x$, there are clearly infinite solutions.
Example
$y =1 \Longrightarrow x = -1$ So the point $( -1 ,1)$ is a solution to the system of equations.
$y =2 \Longrightarrow x = -3$ So the point $( -3 ,2)$ is a solution to the system of equations.
Clearly the system has infinite solutions, but before that any arbitrary point like the point $(2 ,1)$, is not a solution to the system.

Solution:
The two lines represent the given system of equations.
We also note that there is a point of intersection of the both lines. The point is
(2, 3) that is x = 2, y = 3. It belongs to each of the lines and satisfies the both equations.
Therefore, the point (2, 3) is the only solution.

Answer: The system has a unique solution.

Solution:
The both lines are coincident. They have infinite number of points in common that are solutions to the both equations.

Answer: The system has infinite number of solutions.

Solution:
The lines are parallel. They do not have any points in common.
Therefore, there are no points (x, y) that satisfy the both equations simultaneously.

Answer: The system has no solutions.

Solution:
The red and the blue lines pass through two points that are highlighted in the image.
These points must satisfy each equation in the system. Let us review option B)
The first equation is valid when we substitute x and y by points $(-2, 0)$ and $(10 ,8)$, so the equation $6\mathbf{y} -4\mathbf{x} =8$ represents the red line of the image.
Similarly the second equation is fulfilled by substituting by the points $(6 ,0)$ and $(10 ,8)$, so the equation represents the blue line of the image.
The remaining options A, C, D do not satisfy the points on the graph.

Solution:
We solve the both equations for "y = " ( or x = )
$y =4 -4x$ $3x -2 = -\frac{1}{2}y$ $ \Longrightarrow y =4 -6x$ Let us graph the line $\mathbf{y} =4 -4x$ by choosing some arbitrary values for x:
$x = -1 \Longrightarrow y = 8$
$x = 1 \Longrightarrow y = 0$
Having these two points we draw the blue line.

We have to graph the second equation $y =4 -6x$:
$x = -1 \Longrightarrow y =10$
$x =1 \Longrightarrow y = -2$

This line is in red in the image.
We note that the point of intersection of the both line is $(0, 4)$. This point is the solution of the system of linear equations.

Solution:
Let's take the first equation and express the value of x in terms of y.
$x +y =3$ $ \Longrightarrow x =3 -y$

The value we get is substituted into the other equation.

$2x -y =0$ $ \Longrightarrow 2(3 -y) -y =0$ $ \Longrightarrow 6 -2y -y =0$

$6 -3y =0$ $ \Longrightarrow 6 =3y$ $ \Longrightarrow y =2$
Let us get back to the first equation to find x.
$x =3 -y$ $ \Longrightarrow x =3 -2 =1$
The answer is: $x =1\qquad y =2 $

Solution:
We get the variable $\mathbf{y}$ in the first equation.
$ -5y =10x$ $ \Longrightarrow y = -2x$ now we substitute it into the second equation.
$21x -7( -2x) =28$
We can get the value of $x$
$21x +14x =28$ $ \Longrightarrow 35x =28$ $ \Longrightarrow x =\frac{28}{35} =\frac{4}{5}$
To find the value of y we use any of the previous equations.
$y = -2x$
$ \Longrightarrow y = -2(\frac{4}{5}) = -\frac{8}{5}$
The answer is $(\frac{4}{5} , -\frac{8}{5})$

Solution:
We'll first solve each equation for "x ="
$5x -\frac{1}{2}y = -1$
$ \Longrightarrow 5x = -1 +\frac{1}{2}y$ $ \Longrightarrow x = -\frac{1}{5} +\frac{1}{10}y$

$3x -2y =1$ $ \Longrightarrow 3x =1 +2y$ $ \Longrightarrow x =\frac{1}{3} +\frac{2}{3}y$
We get

$ -\frac{1}{5} +\frac{1}{10}y =\frac{1}{3} +\frac{2}{3}y$
We can find the value of y
$\frac{1}{10}y -\frac{2}{3}y =\frac{1}{3} +\frac{1}{5}$ $ \Longrightarrow -\frac{17}{30}y =\frac{8}{15}$

$y = -\frac{16}{17}$
We can find x.
$x = -\frac{1}{5} +\frac{1}{10}y$ $ \Longrightarrow x = -\frac{1}{5} +\frac{1}{10}( -\frac{16}{17})$

$x = -\frac{1}{5} -\frac{16}{170} = -\frac{1}{5} -\frac{8}{85} = -\frac{25}{85} = -\frac{5}{17}$
Thus, the solution of the system of equations is
$x = -\frac{5}{17}\qquad y = -\frac{16}{17}$

Solution:
We solve the both equations for "y = ".

$5x +1 =\frac{1}{2}y$
$ \Longrightarrow y =10x +2$

$3x -1 =2y$
$ \Longrightarrow y =\frac{3x -1}{2}$
Match the both expressions
$10x +2 =\frac{3x -1}{2}$ and find x
$20x +4 =3x -1$
$ \Longrightarrow 17x = -5$
$ \Longrightarrow x =\frac{ -5}{17}$ and we find y.

$y =10(\frac{ -5}{17}) +2 =\frac{-50}{17} +2 =\frac{ -16}{17}$
Answer:
$x =\frac{ -5}{17}$ $y =\frac{-16}{17}$

Solution:
Multiply the first equation by -3 and the second by 2.
$\begin{array}{c} -6x -9y =3 \\ 6x +8y =0\end{array}$
Add the both equations.

$ -6x -9y +6x +8y =3 +0$ $ \Longrightarrow -y =3$ $ \Longrightarrow y = -3$
Substitute this value into the first equation.
$2x +3( -3) = -1$ $ \Longrightarrow 2x =9 -1$ $ \Longrightarrow x =4$
So the solution of the system is x = 4, y = -3

Solution:
To add the equations and one of the two unknowns disappears, the coefficients of this unknown must be the same but with a different sign.
For this we multiply the first equation by $-\frac{1}{2}$ and the second equation by $\frac{1}{3}$

$\begin{array}{c} -\frac{1}{2}(\frac{1}{3}x +\frac{1}{5}y =\frac{2}{7}) \\ \frac{1}{3}(\frac{1}{2}x +\frac{1}{10}y =\frac{3}{7})\end{array}$

$ \Longrightarrow \begin{array}{c} -\frac{1}{6}x -\frac{1}{10}y = -\frac{1}{7} \\ \frac{1}{6}x +\frac{1}{30}y =\frac{1}{7}\end{array}$
This allows us time to add the two equations and eliminate the variable x.

$ -\frac{1}{10}y +\frac{1}{30}y =0$
$ \Longrightarrow \frac{ -2}{30}y =0$
$ \Longrightarrow y =0$
and by substituting this value into the first equation we get
$\frac{1}{3}x =\frac{2}{7}$
$ \Longrightarrow x =\frac{6}{7}$

The answer is
$x =\frac{6}{7}\qquad y =0 $

Solution:
At the intersection point both functions have the same value:
[tex]\frac{1}{2}x-5=6-\frac{7}{8}x[/tex]
[tex]\frac{1}{2}x+\frac{7}{8}x=11 \Rightarrow \frac{11}{8}x=11 \Rightarrow x=8[/tex]
[tex]y=\frac{1}{2}\times 8-5=-1[/tex]
The intersection point is [tex](8, -1)[/tex].

Solution:
The bisector of the second and fourth quadrants is the line [tex]y=-x[/tex].
[tex]\begin{cases} y=-x \\ 3x-2y=10 \end{cases} \Rightarrow 3x+2x=10 \Rightarrow x=2,\ y=-2[/tex]
The intersection point is [tex](2, -2)[/tex].

Solution:
There is no need to find [tex]x[/tex] and [tex]y[/tex] separately. Add the two equations:
[tex](x+4y)+(y+4x)=17+18 \Rightarrow 5x+5y=35 \Rightarrow x+y=7[/tex]

Solution:
Add all three equations — each letter appears twice:
[tex]2(a+b+c)=9+6+7=22 \Rightarrow a+b+c=11[/tex]
[tex]\frac{a+b+c}{3}=\frac{11}{3}=3\frac{2}{3}[/tex]

Solution:
The system has infinitely many solutions when the second equation is the first one multiplied by a number, i.e. the coefficients are proportional:
[tex]\frac{6}{3}=\frac{3a}{-2}=\frac{14}{7}=2[/tex]
[tex]3a=-4 \Rightarrow a=-\frac{4}{3}[/tex]
Then the second equation is [tex]6x-4y=14[/tex], which is the first equation multiplied by [tex]2[/tex].

Solution:
First find the intersection point of the two lines without a parameter:
[tex]2x-1=5-x \Rightarrow 3x=6 \Rightarrow x=2,\ y=3[/tex]
The third line must pass through the point [tex](2, 3)[/tex]: [tex]3=2a-9 \Rightarrow a=6[/tex].

Solution:
Substitute the coordinates of the two points into the equation:
[tex]\begin{cases} 5a-3b=1 \\ -a+b=1 \end{cases}[/tex]
From the second equation [tex]b=a+1[/tex]: [tex]5a-3a-3=1 \Rightarrow 2a=4 \Rightarrow a=2[/tex], [tex]b=3[/tex].
[tex]a-b=2-3=-1[/tex]

Solution:
Clear the denominators in each equation:
[tex]x+y=10x-5y-15 \Rightarrow 9x-6y=15 \Rightarrow 3x-2y=5[/tex]
[tex]7(2x-y)=4(x+2y) \Rightarrow 14x-7y=4x+8y \Rightarrow 10x=15y \Rightarrow x=\frac{3}{2}y[/tex]
Substitute into the first equation: [tex]\frac{9}{2}y-2y=5 \Rightarrow \frac{5}{2}y=5 \Rightarrow y=2[/tex], [tex]x=3[/tex].
The solution of the system is [tex](3, 2)[/tex].

Solution:
Let the numbers we are searching for be x and y.

$\begin{array}{c}2(x +y) =32 \\ x -y =0\end{array}$

We will solve them by substitution.
From the second equation we get $\mathbf{x} =\mathbf{y}$
Now we substitute into the first equation.
$2(x +x) =32$
$ \Longrightarrow 4x =32$
$ \Longrightarrow x =8$
Getting back the second equation we have $\mathbf{x} =\mathbf{y} =8$

Solution:
Let the numbers be x, y.
$\begin{array}{c}x +y =0 \\ x +123 =2y\end{array}$
From the first equation we get
$x = -y$
Now we substitute into the second equation $ -y +123 =2y$ $ \Longrightarrow 123 =3y$ $\Longrightarrow y =\frac{123}{3} =41$ and remembering that $x = -y$ we get $x = -41$ The two numbers we are searching for are $\begin{array}{c}x = -41 \\ y =41\end{array}$

Solution:
Let [tex]x[/tex] be the larger and [tex]y[/tex] the smaller number:
[tex]\begin{cases} x-y=12 \\ x-6=3y \end{cases}[/tex]
From the first equation [tex]x=y+12[/tex]: [tex]y+12-6=3y \Rightarrow 2y=6 \Rightarrow y=3[/tex], [tex]x=15[/tex].
The numbers are [tex]15[/tex] and [tex]3[/tex]. Check: [tex]15-6=9=3\times 3[/tex].

Solution:
Let P be the number of pigs and T the number of turkeys.
$\begin{array}{c}P +T =58 \\ 4P +2T =168\end{array}$
We will use the elimination method to solve it, multiplying the first equation by (-2).
$\begin{array}{c} -2P -2T = -116 \\ 4P +2T =168\end{array}$
Add the both equations and get
$2P =52$ $ \Longrightarrow P =26$
then
$P +T =58$ $ \Longrightarrow 26 +T =58$ $ \Longrightarrow T =32$

Solution:
Let us denote by
J: The money John has.
P: The money Peter has.

Interpreting the problem data we get the following equations:

$\begin{array}{c}J =2P \\ J -6 =P +6\end{array}$
is a system of two linear equations with two variables.
Substitute J in the second equation.
$J -6 =P +6$
$ \Longrightarrow 2P -6 =P +6$
$2P -P =6 +6$
$P =12$ and now we substitute into the first equation
$J =2P =24$
In this way we conclude that John has \$24 and Peter has \$12.

Solution:
Let [tex]t[/tex] be the price of the ticket and [tex]s[/tex] the reservation fee:
[tex]\begin{cases} t+s=108 \\ t=s+102 \end{cases}[/tex]
Substitute [tex]t[/tex] in the first equation: [tex]s+102+s=108 \Rightarrow 2s=6 \Rightarrow s=3[/tex]
The reservation fee is [tex]3[/tex] dollars and the ticket costs [tex]105[/tex] dollars. The quick guess [tex]6[/tex] is wrong: then the ticket would cost [tex]108[/tex] dollars and the total would be [tex]114[/tex] dollars.