First order equation with one unknown variable
$begingroup$
Problem Given: Twelve years ago the age of $A$ was the double of age of $B$ and in the next twelve years, the age of $A$ would be $68$ years less than triple of $B$. Find the current ages.
What I've done is set up the equations
$$2x-12=3(x+12)-68+12$$
where $2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
But the answer doesn't match with the solution on the book $A=52$ and $B=32$.
Could someone tell how to solve?
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
Problem Given: Twelve years ago the age of $A$ was the double of age of $B$ and in the next twelve years, the age of $A$ would be $68$ years less than triple of $B$. Find the current ages.
What I've done is set up the equations
$$2x-12=3(x+12)-68+12$$
where $2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
But the answer doesn't match with the solution on the book $A=52$ and $B=32$.
Could someone tell how to solve?
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
Problem Given: Twelve years ago the age of $A$ was the double of age of $B$ and in the next twelve years, the age of $A$ would be $68$ years less than triple of $B$. Find the current ages.
What I've done is set up the equations
$$2x-12=3(x+12)-68+12$$
where $2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
But the answer doesn't match with the solution on the book $A=52$ and $B=32$.
Could someone tell how to solve?
algebra-precalculus
$endgroup$
Problem Given: Twelve years ago the age of $A$ was the double of age of $B$ and in the next twelve years, the age of $A$ would be $68$ years less than triple of $B$. Find the current ages.
What I've done is set up the equations
$$2x-12=3(x+12)-68+12$$
where $2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
But the answer doesn't match with the solution on the book $A=52$ and $B=32$.
Could someone tell how to solve?
algebra-precalculus
algebra-precalculus
edited Dec 16 '18 at 5:22
Eevee Trainer
6,41811237
6,41811237
asked Dec 16 '18 at 2:06
user459663user459663
306
306
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since for some reason no one felt the need to elaborate on where these equations came from, much less OP's error, I'll do it then.
Though a word of import -- To note for future reference, OP, showing your process might help because otherwise no one can really understand where your own errors come from without it. I happened on your error by simply working the problem the same by mere coincidence; these problems can be solved many different ways, so it's merely luck we coincided.
I'm going to start from the beginning.
Twelve years ago the age of A was the double of age of B and in the next twelve years, the age of A would be 68 years less than triple of B. Find the current ages.
Let $A,B$ denote the respective persons' ages. Then, since "Twelve years ago the age of A was the double of age of B," we know
$$A - 12 =2(B - 12) = 2B - 24 $$
Similarly, per "in the next twelve years, the age of A would be 68 years less than triple of B," we know
$$A + 12 = 3(B+12) - 68 = 3B + 36 - 68 = 3B - 32$$
This gives us a system of equations:
$$A - 12 = 2B - 24$$
$$A + 12 = 3B - 32$$
Equivalently, since $A$ seems easiest to solve for,
$$A = 2B - 12 = 3B - 44$$
Thus, we have the equation in $B$, for which we solve:
$$2B - 12 = 3B - 44 ;;; Rightarrow ;;; B = 32$$
Plug $B = 32$ into any equation of our system of equations, and you'll find $A = 52$.
As for where your error lies, OP, it's this:
What I've done is
$$2x-12=3(x+12)-68+12$$ where
$2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
That last $+12$ is unnecessary. Yes, it is in $12$ years from the present, but that $12$ years is meant to turn $x$ into $x+12$ (as you already have done in parentheses). Since, at that $12$-year-mark, the "$68$ less than the triple" relation is satisfied, you do not add a further $12$.
$endgroup$
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
add a comment |
$begingroup$
Twelve years ago the age of A was the double of age of B so .........$$A-12 = 2(B - 12)$$
$$2B - A = 12 tag{1}$$
In the next twelve years, the age of A would be 68 years less than triple of B so ......$$3(B + 12) - (A + 12) = 68$$
$$3B - A = 44 tag{2}$$
Subtract $(1)$ from $(2)$:
$$B = 32$$
Then $$A = 52$$
$endgroup$
add a comment |
$begingroup$
The problem with your attempted solution is you are using one unknown $x$ for two people. You must use two unknowns for two people.
You can organize all in table:
$$begin{array}{c|c|c}
&12 text{years ago}&Now& 12 text{year after}\
hline
Adam&A-12&A&A+12\
Beth&B-12&B&B+12\
hline
Conditions&A-12=2(B-12)&&A+12=3(B+12)-68end{array}.$$
Can you solve the system of two equations and find $A$ and $B$?
$endgroup$
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
add a comment |
Your Answer
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
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$begingroup$
Since for some reason no one felt the need to elaborate on where these equations came from, much less OP's error, I'll do it then.
Though a word of import -- To note for future reference, OP, showing your process might help because otherwise no one can really understand where your own errors come from without it. I happened on your error by simply working the problem the same by mere coincidence; these problems can be solved many different ways, so it's merely luck we coincided.
I'm going to start from the beginning.
Twelve years ago the age of A was the double of age of B and in the next twelve years, the age of A would be 68 years less than triple of B. Find the current ages.
Let $A,B$ denote the respective persons' ages. Then, since "Twelve years ago the age of A was the double of age of B," we know
$$A - 12 =2(B - 12) = 2B - 24 $$
Similarly, per "in the next twelve years, the age of A would be 68 years less than triple of B," we know
$$A + 12 = 3(B+12) - 68 = 3B + 36 - 68 = 3B - 32$$
This gives us a system of equations:
$$A - 12 = 2B - 24$$
$$A + 12 = 3B - 32$$
Equivalently, since $A$ seems easiest to solve for,
$$A = 2B - 12 = 3B - 44$$
Thus, we have the equation in $B$, for which we solve:
$$2B - 12 = 3B - 44 ;;; Rightarrow ;;; B = 32$$
Plug $B = 32$ into any equation of our system of equations, and you'll find $A = 52$.
As for where your error lies, OP, it's this:
What I've done is
$$2x-12=3(x+12)-68+12$$ where
$2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
That last $+12$ is unnecessary. Yes, it is in $12$ years from the present, but that $12$ years is meant to turn $x$ into $x+12$ (as you already have done in parentheses). Since, at that $12$-year-mark, the "$68$ less than the triple" relation is satisfied, you do not add a further $12$.
$endgroup$
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
add a comment |
$begingroup$
Since for some reason no one felt the need to elaborate on where these equations came from, much less OP's error, I'll do it then.
Though a word of import -- To note for future reference, OP, showing your process might help because otherwise no one can really understand where your own errors come from without it. I happened on your error by simply working the problem the same by mere coincidence; these problems can be solved many different ways, so it's merely luck we coincided.
I'm going to start from the beginning.
Twelve years ago the age of A was the double of age of B and in the next twelve years, the age of A would be 68 years less than triple of B. Find the current ages.
Let $A,B$ denote the respective persons' ages. Then, since "Twelve years ago the age of A was the double of age of B," we know
$$A - 12 =2(B - 12) = 2B - 24 $$
Similarly, per "in the next twelve years, the age of A would be 68 years less than triple of B," we know
$$A + 12 = 3(B+12) - 68 = 3B + 36 - 68 = 3B - 32$$
This gives us a system of equations:
$$A - 12 = 2B - 24$$
$$A + 12 = 3B - 32$$
Equivalently, since $A$ seems easiest to solve for,
$$A = 2B - 12 = 3B - 44$$
Thus, we have the equation in $B$, for which we solve:
$$2B - 12 = 3B - 44 ;;; Rightarrow ;;; B = 32$$
Plug $B = 32$ into any equation of our system of equations, and you'll find $A = 52$.
As for where your error lies, OP, it's this:
What I've done is
$$2x-12=3(x+12)-68+12$$ where
$2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
That last $+12$ is unnecessary. Yes, it is in $12$ years from the present, but that $12$ years is meant to turn $x$ into $x+12$ (as you already have done in parentheses). Since, at that $12$-year-mark, the "$68$ less than the triple" relation is satisfied, you do not add a further $12$.
$endgroup$
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
add a comment |
$begingroup$
Since for some reason no one felt the need to elaborate on where these equations came from, much less OP's error, I'll do it then.
Though a word of import -- To note for future reference, OP, showing your process might help because otherwise no one can really understand where your own errors come from without it. I happened on your error by simply working the problem the same by mere coincidence; these problems can be solved many different ways, so it's merely luck we coincided.
I'm going to start from the beginning.
Twelve years ago the age of A was the double of age of B and in the next twelve years, the age of A would be 68 years less than triple of B. Find the current ages.
Let $A,B$ denote the respective persons' ages. Then, since "Twelve years ago the age of A was the double of age of B," we know
$$A - 12 =2(B - 12) = 2B - 24 $$
Similarly, per "in the next twelve years, the age of A would be 68 years less than triple of B," we know
$$A + 12 = 3(B+12) - 68 = 3B + 36 - 68 = 3B - 32$$
This gives us a system of equations:
$$A - 12 = 2B - 24$$
$$A + 12 = 3B - 32$$
Equivalently, since $A$ seems easiest to solve for,
$$A = 2B - 12 = 3B - 44$$
Thus, we have the equation in $B$, for which we solve:
$$2B - 12 = 3B - 44 ;;; Rightarrow ;;; B = 32$$
Plug $B = 32$ into any equation of our system of equations, and you'll find $A = 52$.
As for where your error lies, OP, it's this:
What I've done is
$$2x-12=3(x+12)-68+12$$ where
$2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
That last $+12$ is unnecessary. Yes, it is in $12$ years from the present, but that $12$ years is meant to turn $x$ into $x+12$ (as you already have done in parentheses). Since, at that $12$-year-mark, the "$68$ less than the triple" relation is satisfied, you do not add a further $12$.
$endgroup$
Since for some reason no one felt the need to elaborate on where these equations came from, much less OP's error, I'll do it then.
Though a word of import -- To note for future reference, OP, showing your process might help because otherwise no one can really understand where your own errors come from without it. I happened on your error by simply working the problem the same by mere coincidence; these problems can be solved many different ways, so it's merely luck we coincided.
I'm going to start from the beginning.
Twelve years ago the age of A was the double of age of B and in the next twelve years, the age of A would be 68 years less than triple of B. Find the current ages.
Let $A,B$ denote the respective persons' ages. Then, since "Twelve years ago the age of A was the double of age of B," we know
$$A - 12 =2(B - 12) = 2B - 24 $$
Similarly, per "in the next twelve years, the age of A would be 68 years less than triple of B," we know
$$A + 12 = 3(B+12) - 68 = 3B + 36 - 68 = 3B - 32$$
This gives us a system of equations:
$$A - 12 = 2B - 24$$
$$A + 12 = 3B - 32$$
Equivalently, since $A$ seems easiest to solve for,
$$A = 2B - 12 = 3B - 44$$
Thus, we have the equation in $B$, for which we solve:
$$2B - 12 = 3B - 44 ;;; Rightarrow ;;; B = 32$$
Plug $B = 32$ into any equation of our system of equations, and you'll find $A = 52$.
As for where your error lies, OP, it's this:
What I've done is
$$2x-12=3(x+12)-68+12$$ where
$2x-12$ is the age of $A$, $3(x+12)$ is the triple of age of $B$ in the future $(+12)$. And the last $12$ because in the sentence says 'is in the next twelve years.'
That last $+12$ is unnecessary. Yes, it is in $12$ years from the present, but that $12$ years is meant to turn $x$ into $x+12$ (as you already have done in parentheses). Since, at that $12$-year-mark, the "$68$ less than the triple" relation is satisfied, you do not add a further $12$.
answered Dec 16 '18 at 5:13
Eevee TrainerEevee Trainer
6,41811237
6,41811237
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
add a comment |
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
1
1
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
$begingroup$
Thank you for the complete answer ^^
$endgroup$
– user459663
Dec 19 '18 at 0:39
add a comment |
$begingroup$
Twelve years ago the age of A was the double of age of B so .........$$A-12 = 2(B - 12)$$
$$2B - A = 12 tag{1}$$
In the next twelve years, the age of A would be 68 years less than triple of B so ......$$3(B + 12) - (A + 12) = 68$$
$$3B - A = 44 tag{2}$$
Subtract $(1)$ from $(2)$:
$$B = 32$$
Then $$A = 52$$
$endgroup$
add a comment |
$begingroup$
Twelve years ago the age of A was the double of age of B so .........$$A-12 = 2(B - 12)$$
$$2B - A = 12 tag{1}$$
In the next twelve years, the age of A would be 68 years less than triple of B so ......$$3(B + 12) - (A + 12) = 68$$
$$3B - A = 44 tag{2}$$
Subtract $(1)$ from $(2)$:
$$B = 32$$
Then $$A = 52$$
$endgroup$
add a comment |
$begingroup$
Twelve years ago the age of A was the double of age of B so .........$$A-12 = 2(B - 12)$$
$$2B - A = 12 tag{1}$$
In the next twelve years, the age of A would be 68 years less than triple of B so ......$$3(B + 12) - (A + 12) = 68$$
$$3B - A = 44 tag{2}$$
Subtract $(1)$ from $(2)$:
$$B = 32$$
Then $$A = 52$$
$endgroup$
Twelve years ago the age of A was the double of age of B so .........$$A-12 = 2(B - 12)$$
$$2B - A = 12 tag{1}$$
In the next twelve years, the age of A would be 68 years less than triple of B so ......$$3(B + 12) - (A + 12) = 68$$
$$3B - A = 44 tag{2}$$
Subtract $(1)$ from $(2)$:
$$B = 32$$
Then $$A = 52$$
edited Dec 16 '18 at 15:34
answered Dec 16 '18 at 4:09
Phil HPhil H
4,2482312
4,2482312
add a comment |
add a comment |
$begingroup$
The problem with your attempted solution is you are using one unknown $x$ for two people. You must use two unknowns for two people.
You can organize all in table:
$$begin{array}{c|c|c}
&12 text{years ago}&Now& 12 text{year after}\
hline
Adam&A-12&A&A+12\
Beth&B-12&B&B+12\
hline
Conditions&A-12=2(B-12)&&A+12=3(B+12)-68end{array}.$$
Can you solve the system of two equations and find $A$ and $B$?
$endgroup$
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
add a comment |
$begingroup$
The problem with your attempted solution is you are using one unknown $x$ for two people. You must use two unknowns for two people.
You can organize all in table:
$$begin{array}{c|c|c}
&12 text{years ago}&Now& 12 text{year after}\
hline
Adam&A-12&A&A+12\
Beth&B-12&B&B+12\
hline
Conditions&A-12=2(B-12)&&A+12=3(B+12)-68end{array}.$$
Can you solve the system of two equations and find $A$ and $B$?
$endgroup$
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
add a comment |
$begingroup$
The problem with your attempted solution is you are using one unknown $x$ for two people. You must use two unknowns for two people.
You can organize all in table:
$$begin{array}{c|c|c}
&12 text{years ago}&Now& 12 text{year after}\
hline
Adam&A-12&A&A+12\
Beth&B-12&B&B+12\
hline
Conditions&A-12=2(B-12)&&A+12=3(B+12)-68end{array}.$$
Can you solve the system of two equations and find $A$ and $B$?
$endgroup$
The problem with your attempted solution is you are using one unknown $x$ for two people. You must use two unknowns for two people.
You can organize all in table:
$$begin{array}{c|c|c}
&12 text{years ago}&Now& 12 text{year after}\
hline
Adam&A-12&A&A+12\
Beth&B-12&B&B+12\
hline
Conditions&A-12=2(B-12)&&A+12=3(B+12)-68end{array}.$$
Can you solve the system of two equations and find $A$ and $B$?
answered Dec 16 '18 at 16:09
farruhotafarruhota
20.4k2739
20.4k2739
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
add a comment |
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
$begingroup$
Thank you, your answer is very clear.
$endgroup$
– user459663
Dec 19 '18 at 0:38
add a comment |
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