solving system of equations $2x^2 - 3y = 23; 3y^2 - 8x = 59$ [closed]
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I couldn't figure out how to solve this:
$2x^2 - 3y = 23$
$3y^2 - 8x = 59$
algebra-precalculus
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
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closed as off-topic by John B, user10354138, Saad, KReiser, Chinnapparaj R Dec 1 '18 at 5:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, user10354138, Saad, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I couldn't figure out how to solve this:
$2x^2 - 3y = 23$
$3y^2 - 8x = 59$
algebra-precalculus
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
$endgroup$
closed as off-topic by John B, user10354138, Saad, KReiser, Chinnapparaj R Dec 1 '18 at 5:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, user10354138, Saad, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
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– Doug M
Dec 1 '18 at 0:44
add a comment |
$begingroup$
I couldn't figure out how to solve this:
$2x^2 - 3y = 23$
$3y^2 - 8x = 59$
algebra-precalculus
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
$endgroup$
I couldn't figure out how to solve this:
$2x^2 - 3y = 23$
$3y^2 - 8x = 59$
algebra-precalculus
algebra-precalculus
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
edited Dec 1 '18 at 0:39
Eray Xx
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
asked Dec 1 '18 at 0:33
Eray XxEray Xx
85
85
closed as off-topic by John B, user10354138, Saad, KReiser, Chinnapparaj R Dec 1 '18 at 5:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, user10354138, Saad, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John B, user10354138, Saad, KReiser, Chinnapparaj R Dec 1 '18 at 5:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, user10354138, Saad, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
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– Doug M
Dec 1 '18 at 0:44
add a comment |
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
$endgroup$
– Doug M
Dec 1 '18 at 0:44
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
$endgroup$
– Doug M
Dec 1 '18 at 0:44
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
$endgroup$
– Doug M
Dec 1 '18 at 0:44
add a comment |
1 Answer
1
active
oldest
votes
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From the first, $3y = (2x^2-23)$. Substituting in the second, we get
$$(2x^2-23)^2 - 24 x = 177$$
Simplifying,
begin{align*}
4x^4 - 92 x^2 + 529 - 24x - 177 &= 0\
x^4 - 23x^2 - 6x + 88 = 0
end{align*}
It is easy to see that 2 and $-4$ are two of the roots. We can now factorize, find the other two roots and substitute in the other equation to find the corresponding values of $y$.
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From the first, $3y = (2x^2-23)$. Substituting in the second, we get
$$(2x^2-23)^2 - 24 x = 177$$
Simplifying,
begin{align*}
4x^4 - 92 x^2 + 529 - 24x - 177 &= 0\
x^4 - 23x^2 - 6x + 88 = 0
end{align*}
It is easy to see that 2 and $-4$ are two of the roots. We can now factorize, find the other two roots and substitute in the other equation to find the corresponding values of $y$.
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
$endgroup$
add a comment |
$begingroup$
From the first, $3y = (2x^2-23)$. Substituting in the second, we get
$$(2x^2-23)^2 - 24 x = 177$$
Simplifying,
begin{align*}
4x^4 - 92 x^2 + 529 - 24x - 177 &= 0\
x^4 - 23x^2 - 6x + 88 = 0
end{align*}
It is easy to see that 2 and $-4$ are two of the roots. We can now factorize, find the other two roots and substitute in the other equation to find the corresponding values of $y$.
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
$endgroup$
add a comment |
$begingroup$
From the first, $3y = (2x^2-23)$. Substituting in the second, we get
$$(2x^2-23)^2 - 24 x = 177$$
Simplifying,
begin{align*}
4x^4 - 92 x^2 + 529 - 24x - 177 &= 0\
x^4 - 23x^2 - 6x + 88 = 0
end{align*}
It is easy to see that 2 and $-4$ are two of the roots. We can now factorize, find the other two roots and substitute in the other equation to find the corresponding values of $y$.
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
$endgroup$
From the first, $3y = (2x^2-23)$. Substituting in the second, we get
$$(2x^2-23)^2 - 24 x = 177$$
Simplifying,
begin{align*}
4x^4 - 92 x^2 + 529 - 24x - 177 &= 0\
x^4 - 23x^2 - 6x + 88 = 0
end{align*}
It is easy to see that 2 and $-4$ are two of the roots. We can now factorize, find the other two roots and substitute in the other equation to find the corresponding values of $y$.
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
answered Dec 1 '18 at 0:45
MuralidharanMuralidharan
43526
43526
add a comment |
add a comment |
$begingroup$
Substitution? $y = frac {2x^2 - 23}{3}$ You will get a quartic equation. If someone was nice, they gave you rational roots. If not, it might be a challenge to factor.
$endgroup$
– Doug M
Dec 1 '18 at 0:44