How to solve this simple equation $frac{46}{y} + y = 25$?
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
add a comment |
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
1
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
1
What aobut $y=23$?
– Yanko
2 days ago
add a comment |
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
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 2 days ago
6005
35.7k751125
35.7k751125
asked 2 days ago
brilliantbrilliant
30729
30729
1
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
1
What aobut $y=23$?
– Yanko
2 days ago
add a comment |
1
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
1
What aobut $y=23$?
– Yanko
2 days ago
1
1
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
1
1
What aobut $y=23$?
– Yanko
2 days ago
What aobut $y=23$?
– Yanko
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$$frac{46}{y}+y = 25$$
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:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
add a comment |
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
add a comment |
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.
add a comment |
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 $y = frac{25-21}{2} = 2$.
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
$$frac{46}{y}+y = 25$$
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:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
add a comment |
$$frac{46}{y}+y = 25$$
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:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
add a comment |
$$frac{46}{y}+y = 25$$
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:
$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$
$$y = 23; quad y = 2$$
$$frac{46}{y}+y = 25$$
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 2 days ago
answered 2 days ago
KM101KM101
5,5511423
5,5511423
add a comment |
add a comment |
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
add a comment |
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
add a comment |
$$y^2-25y+46=0implies y=2,23$$
you can use middle term method or the quadratic formula.
the formula is...
if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$
$$y^2-25y+46=0implies y=2,23$$
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 2 days ago
answered 2 days ago
Rakibul Islam PrinceRakibul Islam Prince
1,035211
1,035211
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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 2 days ago
60056005
35.7k751125
35.7k751125
add a comment |
add a comment |
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 $y = frac{25-21}{2} = 2$.
add a comment |
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 $y = frac{25-21}{2} = 2$.
add a comment |
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 $y = frac{25-21}{2} = 2$.
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 2 days ago
Jonas De SchouwerJonas De Schouwer
1284
1284
add a comment |
add a comment |
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1
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
– Wuestenfux
2 days ago
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
– Thomas Shelby
2 days ago
$2$ is not the only solution - the quadratic formula will give you another.
– Mark Bennet
2 days ago
1
What aobut $y=23$?
– Yanko
2 days ago