Mixing
If one number is thrice the other and their sum is $16$, find the numbers
$begingroup$
If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
begin{align}
x&=3y &iff x-3y=0 &&(1)\
x&=16-3y&&&(2)
end{align}
$$
algebra-precalculus
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
$endgroup$
add a comment |
$begingroup$
If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
begin{align}
x&=3y &iff x-3y=0 &&(1)\
x&=16-3y&&&(2)
end{align}
$$
algebra-precalculus
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
$endgroup$
$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01
add a comment |
$begingroup$
If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
begin{align}
x&=3y &iff x-3y=0 &&(1)\
x&=16-3y&&&(2)
end{align}
$$
algebra-precalculus
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
$endgroup$
If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
begin{align}
x&=3y &iff x-3y=0 &&(1)\
x&=16-3y&&&(2)
end{align}
$$
algebra-precalculus
algebra-precalculus
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
edited Jul 27 '14 at 20:49
user147263
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
asked Jul 27 '14 at 19:57
AbhishekstudentAbhishekstudent
61821231
61821231
$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01
add a comment |
$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01
$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The problem statement says
$$x+3x=16,$$
hence
$$x=4,\3x=12.$$
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
I assume you have $$x=color{red}{3y}, ~~x+y=16$$ Then $3(x+color{red}{y})=3times 16=48$ and so $3x+color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
answered Jul 27 '14 at 20:03
mrsmrs
1
$endgroup$
add a comment |
$begingroup$
$$x=3y$$
$$y+x=16 Rightarrow y+3y=16 Rightarrow 4y=16 Rightarrow y=4$$
So, $x=12$.
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
$endgroup$
add a comment |
$begingroup$
Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.
edited Jan 24 at 2:36
YiFan
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
$endgroup$
add a comment |
$begingroup$
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ tag{1}
x+y=16
$$
$$
tag{2}
x=3y
$$
So, $x-3y=0 tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$tag{1} 3x+3y=48$$
$$tag{2} x-3y=0$$
$$tag{1) + (2}4x=48$$
$$tag{3}x=12$$
Putting in equation $(1)$:
$$tag{1} x+y=16$$
$$tag{1),(3} 12+y=16$$
$$tag{4}y=16-12$$
$$tag{5}y=4$$
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
$endgroup$
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem statement says
$$x+3x=16,$$
hence
$$x=4,\3x=12.$$
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
The problem statement says
$$x+3x=16,$$
hence
$$x=4,\3x=12.$$
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
The problem statement says
$$x+3x=16,$$
hence
$$x=4,\3x=12.$$
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
$endgroup$
The problem statement says
$$x+3x=16,$$
hence
$$x=4,\3x=12.$$
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
answered Jul 27 '14 at 20:06
Yves DaoustYves Daoust
130k676229
130k676229
add a comment |
add a comment |
$begingroup$
I assume you have $$x=color{red}{3y}, ~~x+y=16$$ Then $3(x+color{red}{y})=3times 16=48$ and so $3x+color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
answered Jul 27 '14 at 20:03
mrsmrs
1
$endgroup$
add a comment |
$begingroup$
I assume you have $$x=color{red}{3y}, ~~x+y=16$$ Then $3(x+color{red}{y})=3times 16=48$ and so $3x+color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
answered Jul 27 '14 at 20:03
mrsmrs
1
$endgroup$
add a comment |
$begingroup$
I assume you have $$x=color{red}{3y}, ~~x+y=16$$ Then $3(x+color{red}{y})=3times 16=48$ and so $3x+color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
answered Jul 27 '14 at 20:03
mrsmrs
1
$endgroup$
I assume you have $$x=color{red}{3y}, ~~x+y=16$$ Then $3(x+color{red}{y})=3times 16=48$ and so $3x+color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
answered Jul 27 '14 at 20:03
mrsmrs
1
answered Jul 27 '14 at 20:03
mrsmrs
1
answered Jul 27 '14 at 20:03
mrsmrs
1
answered Jul 27 '14 at 20:03
mrsmrs
1
1
add a comment |
add a comment |
$begingroup$
$$x=3y$$
$$y+x=16 Rightarrow y+3y=16 Rightarrow 4y=16 Rightarrow y=4$$
So, $x=12$.
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
$endgroup$
add a comment |
$begingroup$
$$x=3y$$
$$y+x=16 Rightarrow y+3y=16 Rightarrow 4y=16 Rightarrow y=4$$
So, $x=12$.
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
$endgroup$
add a comment |
$begingroup$
$$x=3y$$
$$y+x=16 Rightarrow y+3y=16 Rightarrow 4y=16 Rightarrow y=4$$
So, $x=12$.
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
$endgroup$
$$x=3y$$
$$y+x=16 Rightarrow y+3y=16 Rightarrow 4y=16 Rightarrow y=4$$
So, $x=12$.
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
answered Jul 27 '14 at 20:04
evindaevinda
4,10031854
4,10031854
add a comment |
add a comment |
$begingroup$
Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.
edited Jan 24 at 2:36
YiFan
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
$endgroup$
add a comment |
$begingroup$
Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.
edited Jan 24 at 2:36
YiFan
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
$endgroup$
add a comment |
$begingroup$
Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.
edited Jan 24 at 2:36
YiFan
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
$endgroup$
Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.
edited Jan 24 at 2:36
YiFan
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
edited Jan 24 at 2:36
YiFan
4,7261727
edited Jan 24 at 2:36
YiFan
4,7261727
edited Jan 24 at 2:36
YiFan
4,7261727
4,7261727
answered Jan 24 at 1:10
user637720user637720
1
answered Jan 24 at 1:10
user637720user637720
1
answered Jan 24 at 1:10
user637720user637720
1
1
add a comment |
add a comment |
$begingroup$
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ tag{1}
x+y=16
$$
$$
tag{2}
x=3y
$$
So, $x-3y=0 tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$tag{1} 3x+3y=48$$
$$tag{2} x-3y=0$$
$$tag{1) + (2}4x=48$$
$$tag{3}x=12$$
Putting in equation $(1)$:
$$tag{1} x+y=16$$
$$tag{1),(3} 12+y=16$$
$$tag{4}y=16-12$$
$$tag{5}y=4$$
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
$endgroup$
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
add a comment |
$begingroup$
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ tag{1}
x+y=16
$$
$$
tag{2}
x=3y
$$
So, $x-3y=0 tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$tag{1} 3x+3y=48$$
$$tag{2} x-3y=0$$
$$tag{1) + (2}4x=48$$
$$tag{3}x=12$$
Putting in equation $(1)$:
$$tag{1} x+y=16$$
$$tag{1),(3} 12+y=16$$
$$tag{4}y=16-12$$
$$tag{5}y=4$$
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
$endgroup$
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
add a comment |
$begingroup$
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ tag{1}
x+y=16
$$
$$
tag{2}
x=3y
$$
So, $x-3y=0 tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$tag{1} 3x+3y=48$$
$$tag{2} x-3y=0$$
$$tag{1) + (2}4x=48$$
$$tag{3}x=12$$
Putting in equation $(1)$:
$$tag{1} x+y=16$$
$$tag{1),(3} 12+y=16$$
$$tag{4}y=16-12$$
$$tag{5}y=4$$
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
$endgroup$
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ tag{1}
x+y=16
$$
$$
tag{2}
x=3y
$$
So, $x-3y=0 tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$tag{1} 3x+3y=48$$
$$tag{2} x-3y=0$$
$$tag{1) + (2}4x=48$$
$$tag{3}x=12$$
Putting in equation $(1)$:
$$tag{1} x+y=16$$
$$tag{1),(3} 12+y=16$$
$$tag{4}y=16-12$$
$$tag{5}y=4$$
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
edited Jan 24 at 2:47
Adam Hrankowski
2,094930
2,094930
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
answered May 3 '18 at 15:50
Rachit GuptaRachit Gupta
21
21
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
add a comment |
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
1
1
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
$begingroup$
Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers.
$endgroup$
– Bill O'Haran
May 3 '18 at 15:57
add a comment |
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$begingroup$
One number is $x$, the other $3x$. They add to 16, so ... .
$endgroup$
– Chris Leary
Jul 27 '14 at 20:01
$begingroup$
the second part of the questions states "their sum is 16". this can be expressed as x+y=16
$endgroup$
– Mufasa
Jul 27 '14 at 20:01