Mixing
Help with possible algebra equation
2
$begingroup$
My friend has sent me this puzzle, but I’m terrible with Algebra, could someone please explain how to solve it step by step.
For example, how would I find the value of Q?
Puzzle
I wasn’t able to post the picture as my reputation is too low sorry!
algebra-precalculus puzzle
asked Jan 12 at 11:02
NickNick
132
$endgroup$
add a comment |
2
$begingroup$
My friend has sent me this puzzle, but I’m terrible with Algebra, could someone please explain how to solve it step by step.
For example, how would I find the value of Q?
Puzzle
I wasn’t able to post the picture as my reputation is too low sorry!
algebra-precalculus puzzle
asked Jan 12 at 11:02
NickNick
132
$endgroup$
add a comment |
2
2
2
$begingroup$
My friend has sent me this puzzle, but I’m terrible with Algebra, could someone please explain how to solve it step by step.
For example, how would I find the value of Q?
Puzzle
I wasn’t able to post the picture as my reputation is too low sorry!
algebra-precalculus puzzle
asked Jan 12 at 11:02
NickNick
132
$endgroup$
My friend has sent me this puzzle, but I’m terrible with Algebra, could someone please explain how to solve it step by step.
For example, how would I find the value of Q?
Puzzle
I wasn’t able to post the picture as my reputation is too low sorry!
algebra-precalculus puzzle
algebra-precalculus puzzle
asked Jan 12 at 11:02
NickNick
132
asked Jan 12 at 11:02
NickNick
132
asked Jan 12 at 11:02
NickNick
132
asked Jan 12 at 11:02
NickNick
132
asked Jan 12 at 11:02
NickNick
132
132
add a comment |
add a comment |
2 Answers
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- In the second row, you have $K+K+K+K = 28$, which means $4K = 28 iff color{blue}{K = 7}$.
- In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields $Q+3(7) = 30 iff color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.
- In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 iff color{purple}{L+2Z = 37} tag{1} $.
- In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 iff color{purple}{Z+L = 24} tag{2}$.
- You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $color{red}{Z = 13}$. Plugging it in in either equation yields $color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.
answered Jan 12 at 11:19
KM101KM101
5,9251524
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1
$begingroup$
It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
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2 Answers
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2 Answers
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$begingroup$
- In the second row, you have $K+K+K+K = 28$, which means $4K = 28 iff color{blue}{K = 7}$.
- In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields $Q+3(7) = 30 iff color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.
- In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 iff color{purple}{L+2Z = 37} tag{1} $.
- In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 iff color{purple}{Z+L = 24} tag{2}$.
- You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $color{red}{Z = 13}$. Plugging it in in either equation yields $color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.
answered Jan 12 at 11:19
KM101KM101
5,9251524
$endgroup$
add a comment |
0
$begingroup$
- In the second row, you have $K+K+K+K = 28$, which means $4K = 28 iff color{blue}{K = 7}$.
- In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields $Q+3(7) = 30 iff color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.
- In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 iff color{purple}{L+2Z = 37} tag{1} $.
- In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 iff color{purple}{Z+L = 24} tag{2}$.
- You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $color{red}{Z = 13}$. Plugging it in in either equation yields $color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.
answered Jan 12 at 11:19
KM101KM101
5,9251524
$endgroup$
add a comment |
0
0
0
$begingroup$
- In the second row, you have $K+K+K+K = 28$, which means $4K = 28 iff color{blue}{K = 7}$.
- In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields $Q+3(7) = 30 iff color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.
- In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 iff color{purple}{L+2Z = 37} tag{1} $.
- In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 iff color{purple}{Z+L = 24} tag{2}$.
- You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $color{red}{Z = 13}$. Plugging it in in either equation yields $color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.
answered Jan 12 at 11:19
KM101KM101
5,9251524
$endgroup$
- In the second row, you have $K+K+K+K = 28$, which means $4K = 28 iff color{blue}{K = 7}$.
- In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields $Q+3(7) = 30 iff color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.
- In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 iff color{purple}{L+2Z = 37} tag{1} $.
- In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 iff color{purple}{Z+L = 24} tag{2}$.
- You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $color{red}{Z = 13}$. Plugging it in in either equation yields $color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.
answered Jan 12 at 11:19
KM101KM101
5,9251524
answered Jan 12 at 11:19
KM101KM101
5,9251524
answered Jan 12 at 11:19
KM101KM101
5,9251524
answered Jan 12 at 11:19
KM101KM101
5,9251524
5,9251524
add a comment |
add a comment |
1
$begingroup$
It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
1
$begingroup$
It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
1
1
1
$begingroup$
It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
$endgroup$
It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
answered Jan 12 at 11:08
Yves DaoustYves Daoust
127k673226
127k673226
add a comment |
add a comment |
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