Let
x--------> the number of blue beads
y--------> the number of red beads
we know that
[tex] x+y=49 [/tex]
[tex] x=49-y [/tex] -------> equation [tex] 1 [/tex]
[tex] x=6y [/tex] ------> equation [tex] 2 [/tex]
equate equation [tex] 1 [/tex] and equation [tex] 2 [/tex]
[tex] 49-y=6y\\ 6y+y=49\\ 7y=49\\\\ y=\frac{49}{7} \\ \\ y=7 [/tex]
find the value of x
[tex] x=6*7\\ x=42 [/tex]
therefore
the answer is
Ivan has [tex] 42 [/tex] [tex] blue beads [/tex]
The total number of blue beads with Ivan is [tex]\boxed{\bf 42}[/tex].
Further explanation:
It is given that Ivan has [tex]6[/tex] times as many blue beads as red beads.
The total number of beads are [tex]49[/tex].
Calculation:
Assume the beads of red color are denoted by [tex]R[/tex] and the beads of blue color are denoted by [tex]B[/tex].
Now, given that there are total [tex]49[/tex] beads and this can be written in the form of an equation as follows:
[tex]\boxed{R+B=49}[/tex] ......(1)
Also, given that Ivan has [tex]6[/tex] times as many blue beads as red beads and this can written as follows:
[tex]\boxed{6R=B}[/tex] ......(2)
Substitute the value [tex]6R=B[/tex] in equation (1), we get
[tex]\begin{aligned}R+6R&=49\\7R&=49\\R&=\dfrac{49}{7}\\R&=7\end{aligned}[/tex]
Therefore, Ivan has [tex]7[/tex] red beads.
Substitute [tex]R=7[/tex] in equation (1).
[tex]\begin{aligned}B&=6\cdot 7\\&=42\end{aligned}[/tex]
This implies that number of blue beads are [tex]42[/tex].
Thus, the total number of blue beads with Ivan is [tex]\boxed{\bf 42}[/tex].
Learn more
1. Problem on the equation of the circle https://brainly.com/question/1952668
2. Problem on the center and radius of an equation https://brainly.com/question/9510228
3. Problem on the general form of the equation of the circle https://brainly.com/question/1506955
Answer details:
Grade: Middle school
Subject: Mathematics
Chapter: Linear equations in two variables
Keywords: Linear equations in one variable, linear equations in two variables, substitution, elimination, function, sets, real numbers, ordinates, abscissa, interval.