Respuesta :
For each of the numbers, let us attribute a different letter for each symbol, then we will solve the system of equations.
1)
Let it be:
Box = b ; Arrow = a ; Owl = o
then, we write the following system of equations:
[tex]\begin{gathered} b\times a=o \\ o-b=45 \\ a+b=19 \end{gathered}[/tex]To solve the present problem, let's start with summing the second and the third equations. This will give us the sum of a and o equal to a number. Then we can isolate a and substitute in the first equation. It will allow us to determine the value of b, as follows:
2nd + 3rd equations:
[tex]\begin{gathered} o+b+a-b=45+19 \\ o+a=64 \\ a=64-o \end{gathered}[/tex]substituting in the first equation, we find:
[tex]\begin{gathered} b\times(64-o)=o \\ b=\frac{o}{64-o} \end{gathered}[/tex]Now we can substitute it into the second equation, as follows:
[tex]\begin{gathered} o-\frac{o}{64-o}=45\Rightarrow\frac{64o-o^{2}-o}{64-o}=45 \\ 63o-o^2=45\times64-45o\Rightarrow63o-o^{2}=2880-45o \\ o^{2}-108o+2880=0 \end{gathered}[/tex]Solving the present equation by using the Bhaskara formula, we find two possible values for o:
[tex]\begin{gathered} o_1=48 \\ o_2=60 \end{gathered}[/tex]Let us solve to each possible:
For o = 48, we substitute into the first two equations and we find the following:
[tex]\begin{gathered} b\times a=48 \\ 48-b=45\Rightarrow b=3 \\ \\ \Rightarrow3\times a=48\Rightarrow a=\frac{48}{3} \\ a=16 \end{gathered}[/tex]One of the possible solutions is:
[tex]\begin{gathered} o=48 \\ a=16 \\ b=3 \end{gathered}[/tex]The second possible solution, for o = 60, we have:
[tex]\begin{gathered} b\times a=60 \\ 60-b=45\Rightarrow b=15 \\ \\ 15\times a=60\Rightarrow a=\frac{60}{15} \\ a=4 \end{gathered}[/tex]Another possible solution is the following:
[tex]\begin{gathered} a=4 \\ b=15 \\ o=60 \end{gathered}[/tex]Now, let's work on the second problem:
2)
Let's set the value of: heart = h, the other unknown figure = u
[tex]\begin{gathered} u\times2=h-6 \\ h-u=23 \end{gathered}[/tex]To solve it, let's isolate h in the second equation and substitute it into the first one to find the value of u.
[tex]\begin{gathered} h=23+u \\ 2u=u+23-6 \\ \Rightarrow u=17 \end{gathered}[/tex]Substituting it in the second equation again, we have:
[tex]\begin{gathered} h-17=23 \\ h=17+23=40 \\ h=40 \end{gathered}[/tex]From the solution developed above, we conclude that the solution is:
[tex]\begin{gathered} h=40 \\ u=17 \end{gathered}[/tex]3)
For a sake of simplicity, let us call the three figures in the first relation of a, b, and c, respectively. Then we have:
[tex]\begin{gathered} a\times b=c \\ \frac{a}{b}=6 \\ c-a=36 \end{gathered}[/tex]By isolating b in the second equation, and c in the third equation, and then substituting both into the first one, we have the following:
[tex]\begin{gathered} b=\frac{a}{6} \\ c=a+36 \\ \\ \Rightarrow a\times\frac{a}{6}=a+36 \\ a^2-6a-216=0 \end{gathered}[/tex]The solutions for the present equation, from the Bhaskara formula, are:
[tex]\begin{gathered} a_1=-12 \\ a_2=18 \end{gathered}[/tex]For the value a = -12, we have:
[tex]\begin{gathered} -12b=c \\ -\frac{12}{b}=6\Rightarrow b=-2 \\ c-(-12)=36\Rightarrow c=24 \end{gathered}[/tex]This would lead to the first solution, which is:
[tex]\begin{gathered} a=-12 \\ b=-2 \\ c=24 \end{gathered}[/tex]For the value a = 18, we would have:
[tex]\begin{gathered} 18b=c \\ \frac{18}{b}=6\Rightarrow b=3 \\ c-18=36\Rightarrow c=54 \end{gathered}[/tex]And the second possible solution for the number 3 is:
[tex]\begin{gathered} a=18 \\ b=3 \\ c=54 \end{gathered}[/tex]Now, let's solve number 4.
4)
Let's name the three figures as a, b, and c, respectively.
[tex]\begin{gathered} \frac{a}{b}=c \\ c\times a=288 \\ a-b=40 \end{gathered}[/tex]The first step is to isolate c in the second equation and b in the third, substitute in the first equation, then we will find the value of a.
[tex]\begin{gathered} c=\frac{288}{a} \\ b=a-40 \\ \\ \frac{a}{a-40}=\frac{288}{a}\Rightarrow a^{2}=288a-11,520 \\ a^{2}-288a+11,520=0 \end{gathered}[/tex]The values which are the solution for this equation, from the Bhaskara equation, are:
[tex]\begin{gathered} a_1=240 \\ a_2=48 \end{gathered}[/tex]Using the value a = 240, we find the following values:
[tex]\begin{gathered} \frac{240}{b}=c \\ 240c=288\Rightarrow c=1.2 \\ 240-b=40\Rightarrow b=200 \end{gathered}[/tex]From this value of a, the solution is:
[tex]\begin{gathered} a=240 \\ b=200 \\ c=1.2 \end{gathered}[/tex]For the second value of a, a = 48, we have:
[tex]\begin{gathered} \frac{48}{b}=c \\ c\times48=288\Rightarrow c=6 \\ 48-b=40\Rightarrow b=8 \end{gathered}[/tex]From this, the last solution for the number 4 is:
[tex]\begin{gathered} a=48 \\ b=8 \\ c=6 \end{gathered}[/tex]