How to solve this simple equation $frac{46}{y} + y = 25$?
$begingroup$
How do I solve this simple equation? $frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
How do I solve this simple equation? $frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
algebra-precalculus
$endgroup$
1
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
1
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22
add a comment |
$begingroup$
How do I solve this simple equation? $frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
algebra-precalculus
$endgroup$
How do I solve this simple equation? $frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
algebra-precalculus
algebra-precalculus
edited Jan 5 at 13:28
6005
37k751127
37k751127
asked Jan 5 at 13:18
brilliantbrilliant
30739
30739
1
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
1
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22
add a comment |
1
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
1
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22
1
1
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
1
1
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$frac{46}{y}+y = 25$$
Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
- $$y-23 = 0 iff y = 23$$
- $$y-2 = 0 iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
$endgroup$
add a comment |
$begingroup$
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
$endgroup$
add a comment |
$begingroup$
Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:
begin{align*}
left(frac{46}{y} + yright) cdot y &= 25 cdot y \
46 + y^2 &= 25y
end{align*}
Now we bring all the terms over to one side:
$$
y^2 - 25y + 46 = 0
$$
so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)
If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.
$endgroup$
add a comment |
$begingroup$
Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = frac{25 pm sqrt{21^2}}{2}$$
So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$frac{46}{y}+y = 25$$
Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
- $$y-23 = 0 iff y = 23$$
- $$y-2 = 0 iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
$endgroup$
add a comment |
$begingroup$
$$frac{46}{y}+y = 25$$
Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
- $$y-23 = 0 iff y = 23$$
- $$y-2 = 0 iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
$endgroup$
add a comment |
$begingroup$
$$frac{46}{y}+y = 25$$
Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
- $$y-23 = 0 iff y = 23$$
- $$y-2 = 0 iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
$endgroup$
$$frac{46}{y}+y = 25$$
Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
- $$y-23 = 0 iff y = 23$$
- $$y-2 = 0 iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
edited Jan 5 at 13:34
answered Jan 5 at 13:27
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
$begingroup$
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
$endgroup$
add a comment |
$begingroup$
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
$endgroup$
add a comment |
$begingroup$
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
$endgroup$
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
edited Jan 5 at 13:30
answered Jan 5 at 13:22
Rakibul Islam PrinceRakibul Islam Prince
988211
988211
add a comment |
add a comment |
$begingroup$
Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:
begin{align*}
left(frac{46}{y} + yright) cdot y &= 25 cdot y \
46 + y^2 &= 25y
end{align*}
Now we bring all the terms over to one side:
$$
y^2 - 25y + 46 = 0
$$
so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)
If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.
$endgroup$
add a comment |
$begingroup$
Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:
begin{align*}
left(frac{46}{y} + yright) cdot y &= 25 cdot y \
46 + y^2 &= 25y
end{align*}
Now we bring all the terms over to one side:
$$
y^2 - 25y + 46 = 0
$$
so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)
If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.
$endgroup$
add a comment |
$begingroup$
Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:
begin{align*}
left(frac{46}{y} + yright) cdot y &= 25 cdot y \
46 + y^2 &= 25y
end{align*}
Now we bring all the terms over to one side:
$$
y^2 - 25y + 46 = 0
$$
so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)
If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.
$endgroup$
Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:
begin{align*}
left(frac{46}{y} + yright) cdot y &= 25 cdot y \
46 + y^2 &= 25y
end{align*}
Now we bring all the terms over to one side:
$$
y^2 - 25y + 46 = 0
$$
so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)
If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.
answered Jan 5 at 13:27
60056005
37k751127
37k751127
add a comment |
add a comment |
$begingroup$
Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = frac{25 pm sqrt{21^2}}{2}$$
So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.
$endgroup$
add a comment |
$begingroup$
Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = frac{25 pm sqrt{21^2}}{2}$$
So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.
$endgroup$
add a comment |
$begingroup$
Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = frac{25 pm sqrt{21^2}}{2}$$
So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.
$endgroup$
Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = frac{25 pm sqrt{21^2}}{2}$$
So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.
answered Jan 5 at 13:29
Jonas De SchouwerJonas De Schouwer
3748
3748
add a comment |
add a comment |
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1
$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19
$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20
$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21
1
$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22