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1 vote

What is this condition?

In order to have a vector $c$ with $Ac=b$, you need $b$ to be in the column space of $A$ (since an expression of the form $Ac$ is a linear combination of the columns of $A$.)
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2 votes
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Finding Solutions to a Symmetric Non-Linear Equation (for Some Cases Beside $x = y = z$)

To simplify the calculations, doing the same as @Dietrich Burde did in his answer, let $a=\frac 1{2022}$ and you end with the cubic $$12 a x^3-36 x^2-3 a^3 x+a^2=0$$ which has three real roots since $$...
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2 votes

Finding Solutions to a Symmetric Non-Linear Equation (for Some Cases Beside $x = y = z$)

Multiplying the first equation by $z$, the second by $y$ and the third by $x$ we obtain a system of polynomial equations \begin{align*} 4044xz - 4044yz - z + 2022 & = 0,\\ - 4044xy + 4044yz - y + ...
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3 votes
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Solve $ \frac{1}{xy} = \frac{x}{yz} + 1, \:\: \frac{1}{yz} = \frac{y}{zx} + 1, \:\: \frac{1}{zx} = \frac{z}{xy} + 1. $

We see that the first equation is equivalent to: $$\frac{1}{xy} = \frac{x}{yz} + 1\ \Rightarrow z = x^{2}+xyz \ \Rightarrow z(1-xy) = x^{2}$$ Since $x^{2}$ and $z$ are positive, we must have $1-xy >...
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1 vote

Time and Work Intuitive way

I'd suggest using conceptually simpler variables, like taking per day work of men as 'm' and that of women as 'w'. Rest of my answer is the solution with these, rather than what you asked. Writing ...
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1 vote
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Time and Work Intuitive way

You have made a calculation error: the correct answer should be $\displaystyle M=\frac{11}{5},\ W=\frac 15$. Thus 10 women work 2 units a day and hence take 40 days to complete the assignment.
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-1 votes

Time and Work Intuitive way

There is a fundamental problem with this- it assumes that every man works at the same rate and every woman works at the same rate- and that is just not true! But assuming this, let x be the rate at ...
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3 votes

How do you solve $a^2b^2c-8abc+c^2+ac+bc-ab+8c+8=0$?

This is by way of an extended comment to show one way of simplifying what you are given. If you set $p=ab, q=a+b$ this becomes $c^2+(p^2-8p+q+8)c+(8-q)=0$ or $c^2+\left((p-4)^2+q-8\right)+8-q=0$ Now ...
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Enumerating solutions of incomplete row reduced echelon form over multivariate polynomial ring over GF2

While this is not strictly the answer I was originally looking for, this how I tackled my main question of enumerating all possible solutions. I used a different represenation of the problem and a ...
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1 vote

Solving algebraic simultaneous equations

As pointed out in the comments, the third equation is a combination of the first two. Since $a_3=-a_1-a_2$ and $b_3=-b_1-b_2$, we have the equivalent system $$a_1=b_1x+(b_1+b_2)z\\a_2=b_2y-b_1x$$ or $$...
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$\frac{x^2}{x} =x$ isn't an equation true for all values of x?

Yes, many mathematical statements (such as equations) are only true under certain conditions (such as restrictions on the values of variables). It's common to express these as if-then statements, as ...
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$\frac{x^2}{x} =x$ isn't an equation true for all values of x?

In this case, the equation $\frac{x^2} x = x$ should be thought of as a condition, selecting for particular values of $x$. In particular, this equation is equivalent to the condition $x \neq 0$. Your ...
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2 votes

Determining whether two given lines intersect or are parallel

If two lines intersect, they must necessarily lie in the same plane, i.e. be coplanar. $$L_1: \vec r= (5,0,-3)-\lambda(6,3,9)$$$$= (5,0,-3)-3\lambda(2,1,3)= (5,0,-3)+\mu(2,1,3)$$ $$L_2: \vec r=(5,-1,...
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0 votes

Determining whether two given lines intersect or are parallel

Um, it might be a dumb method, but just assume that they intersect, and let x, y, z be the coordinates that is the intersection point. so you'll have three sets of s-t equations listed as follows : 5 -...
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2 votes
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Determining whether two given lines intersect or are parallel

For the two to intersect, there must be a solution to the overdetermined system $$5-6t = 5+3s,$$ $$-3t = -1-4s,$$ $$-3-9t = 11+2s.$$ Pick two equations where the coefficients of $s$ and $t$ are not ...
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Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$

Your equation has no non-zero solutions. Lemma. The Diophantine equation $3x^2+y^2=2z^2$ has no non-zero solutions in integers. Proof. Let $x_0$ is smallest positive integer such that there are ...
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1 vote
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Neural Networks: Different Depths and Widths But Same Number of Parameters

Assume all hidden layers have exactly the same size $b_m$. By the quadratic formula applied to your equation with a box, considering $b_m$ should be positive, one gets: $$b_m \approx \hat{b_m} = \...
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2 votes

Is there an efficient way to solve this problem?

An alternative solution, easily generalizable to higher degree polynomials, would go like this: First, consider the polynomial: $P_1(x) = \frac{q_1(x - p_2)(x - p_3)}{(p_1 - p_2)(p_1 - p_3)}$. Then, $...
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1 vote
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Is there an efficient way to solve this problem?

Something that seems overlooked in the present answer and comments is that the question only asked to show that such a curve exists. Nothing was said about getting an explicit formula for any such ...
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1 vote

Is there an efficient way to solve this problem?

I see what you mean. Just to see what the result of the row reduction is I entered the relevant matrix into a computer algebra system (by typing row reduce {{1,p_1,(p_1)^2,q_1},{1,p_2,(p_2)^2,q_2},{1,...
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2 votes

How to solve equations involving reciprocal square roots of quadratics?

Yes! This is quite easy: $$\begin{align*}\frac{2x-8}{\sqrt{2x^2-16x+34}}+\frac{2x-3}{\sqrt{2x^2-6x+5}}&=0\\ \frac{2x-8}{\sqrt{2x^2-16x+34}}&=-\frac{2x-3}{\sqrt{2x^2-6x+5}}\\ \left(\frac{2x-8}{\...
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4 votes
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Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$

This isn't a full answer, but noting that $$2a+2b+2c=6d$$ first subtract this from the second equation to obtain $$(a-1)^2+(b-1)^2+(c-1)^2=4(d-1)^2$$ Then set $A=a-1, B=b-1, C=c-1, D=d-1$ to get the ...
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-1 votes

Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$

$a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$ Just set $(a, b, c)=(k_1d+l_1, k_2d+l_2, k_3d+l_3)$, since if we have quadratic polynomials for $a, b, c$, the 4th degree polynomials have to be in $a^2+b^2+c^2$. $4d^...
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0 votes
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Constant $SU(2)$ ASD connections on $\mathbb R^4$ are Flat

I've worked out a solution, which I'll post in case anyone might find this useful. This is similar to the argument that the ASD connections are critical points for the Yang-Mills functional First ...
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1 vote

How can I find parameter of the rational map?

It is an approximation of the answer using brute force and visual analysis. First analyze the map f find critical points compute Mandelbrot set find parameter $\rho$ : point on the unit circle which ...
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1 vote
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Show three nonlinear equations have finitely many solutions

I guess that you assume that $c_1$, $c_2$ (and not $x_1-c_1$, $x_2-c_2$) are non null, and that the unknowns are $x_1$ and $x_2$? For all real number $x$, set $f(x) = p~\mathrm{sign}(x)~|x|^{p-1} + \...
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1 vote

Solve the trig system of equations

Suggested start too long for a comment. The third and fourth equations have the form $$ x = r\cos \sigma + s \sin \tau $$ $$ y = r\sin \sigma + s \cos \tau $$ with $x$, $y$, $r$ and $s$ known since ...
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4 votes

Solve the trig system of equations

Here is a method. $x^2 + y^2 = $ $d^2(\cos^2\theta + \sin^2\theta) + b^2(\cos^2\phi + \sin^2\phi) - 2db (\cos\theta\cos\phi - \sin\theta\sin\phi) = $ $ d^2 + b^2 - 2db\cos(\theta + \phi)$ You said ...
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0 votes

A question about Quadratic Simultaneous Equations

Another way to interpret your method of solution is that inserting $ \ x \ = \ 2 - y \ $ into the quadratic equation leads to $$ y \ \ = \ \ (2 - y)^2 \ + \ 2·(2 - y) \ + \ 2 \ \ \Rightarrow \ \ y \ \ ...
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How can I tell if this parametric equation intersects?

The answer is yes, they will intersect. Actually, your starting point is correct, but you need to define your problem in a mathematical way. In brief, if there exists a pair $(t,u)$ that makes $x(t)=x(...
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0 votes

Solve simultaneous equations

Rewrite the last equation as $$\lambda_g-\lambda_d= k \big[\lambda_d\big]^{-2/3}$$ Cubing both sides gives a pentic equation in $\big[\lambda_d\big]^{1/3}$ : so, no explicit solution and a numerical ...
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3 votes
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Does the following system of equations have a closed form solution?

Write the equations as $$(A-a)x=p B\tag 1$$ $$(A-b)y=q B\tag 2$$ $$(A-c)z=r B\tag 3$$ $$A=p^2+q^2+r^2\tag 4$$ $$B=px+qy+rz\tag 5$$ Using $(1)$, $(2)$ and $(3)$, we have $$p=\frac{x (A-a)}{B} \qquad \...
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2 votes

$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$

Let $\sqrt[3]{10-x}=a,\sqrt[3]{30-x}=b,\sqrt[3]{15-x}=c, \sqrt[3]{25-x}=d$ We have $a+b=c+d\ \ \ \ (1)$ Again $a^3+b^3=c^3+d^3$ $\iff(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$ So, either case$\#1: a+b=0$ or ...
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5 votes
Accepted

$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$

Suppose $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$ I will use the following \begin{align} A^3=&(p+q)^3 \\ =&p^3+q^3+3pq(p+q) \\ =&p^3+q^3+3pq(A) \end{align} Then \...
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4 votes

How to solve the triangulation problem?

Hint: start with $R_b=z_1+R_a$. Then by rearranging the first two equations, squaring and adding, we can obtain: $R_a^2=(A-1)^2(x^2+y^2)$. Similarly, squaring and adding third and fourth equation, we ...
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2 votes

How to solve the triangulation problem?

From a trigonometric identity, the values of $x$ and $y$ must satisfy these equations simultaneously: $$x^2+y^2 = \frac{R_a^2}{(A-1)^2}$$ $$x^2+y^2 = \frac{R_b^2}{(B-1)^2}$$ $$x^2+y^2 = \frac{R_c^2}{(...
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11 votes
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System of inequations

Rewrite the inequalities like this: $$q+r<s+t<p+q<r+s<t+p$$ From this, we immediately see: $$\begin{array}{c}q<t\\r<p\\q<s\\t<r\\s<p\end{array}$$ (Say, concentrate on the ...
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2 votes
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Cramer's rule and centre of a circle

To be more pedagogical, he should have had the last row be all $1$'s so as to match the variable he states i.e. $R^2 - |z|^2$ as you say, but the end result is the same - by determinant rules, you can ...
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1 vote
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How to solve for m in this equation without using the quadratic formula

Converting a comment to an answer, as requested ... If all you know is that $m_1m_2=2$ (where $m_1$ and $m_2$ are the roots of the quadratic), then you can't find a specific value for $b^2−a^2$. By ...
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A question about Quadratic Simultaneous Equations

The blue line $y=2-x$ cuts the parabola $y=x^2+2x+2$ at $(3,-5)$ and $(0,2)$. But when you put y=5 in the second equation, you are actually solving the parabola with the green line $y=5$, which cuts ...
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3 votes

Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

Squaring both sides works out not so bad. Move the big radical to the other side and square: $$\sqrt{x+3}+\sqrt{5-x} = 2\sqrt{15+2x-x^2}-4$$ $$8 +2\sqrt{15+2x-x^2} = 4(15+2x-x^2)-16\sqrt{15+2x-x^2} +...
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LU decomposition needed for this problem

You must first notice that NOT ALL matrices can be written in LU form. Row changes may be needed, so the correct form for any matrix is that there exists a permutation matrix P such that PA=LU. Now we ...
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Prove that a system $(S)$ is inconsistent

In addition to the good first answer by @Dragonoverlord3000, here is an intuitive one. $A X = B$ means $B$ is a linear combination of the column vectors of $A$. (Or, if you prefer, it is in the vector ...
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1 vote
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Prove that a system $(S)$ is inconsistent

The condition $(a_{1k}, a_{2k}, \dots, a_{mk})B = 0$ is equivalent to $A^TB = 0$, as stated in the comments by Sammy Black. But then we can multiply by $A^T$ on both sides of $AX = B$ giving us $A^TAX ...
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1 vote
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LU decomposition needed for this problem

Gaussian elimination is performed on the augmented matrix $[A | b]$, that is $b$ must be known. LU is a factorization method that performs Gaussian elimination on $A$ (no $b$). You can then solve the ...
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2 votes

Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

Using dxiv's hint. $a^2+b^2 = (a+b)^2 - 2ab\\ 8 = (a+b)^2 - 2ab\\ -2ab = 8 - (a+b)^2$ Substitute into $a+b - 2ab = -4\\ (a+b) + 8 - (a+b)^2 = -4\\ (a+b)^2 -(a+b) - 12 = 0\\ (a+b - 4)(a+b + 3) = 0$ $a+...
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5 votes
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Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

Squaring both sides was too complicated so I tried to substitute $a = \sqrt{5−x}, ~b = \sqrt{x + 3}.$ This gave the new multivariable equation: $~a+b−2ab=−4$. $a + b - 2ab = -4.$ $a^2 + b^2 = 8.$ ...
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2 votes

Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$?

Taking the resultant of $p=a^3-33ab^2+217$ and $q=3a^2b-11b^3-18$ with respect to $b$ immediately gives $$ (4a^4 - 14a^3 + 111a^2 + 217a + 961)(4a^2 + 14a + 49)(2a^2 + 7a - 31)(2a - 7)=0. $$ Hence the ...
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3 votes

Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$?

This is one of the other approaches you would like to try. Multiply the first equation by $18$, multiply the second one by $217$, then add them together, we get: $18a^3-18*33ab^2+217*3a^2b-217*11b^3=0$...
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5 votes

Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$?

Does this help? $$(a-\sqrt{11} i b)^3 = a^3-3\sqrt{11}i a^2 b -33ab^2+11\sqrt{11}ib^3$$ $$=(a^3 -33ab^2)-\sqrt{11}i(3 a^2 b-11b^3)$$ $$=-217-\sqrt{11} \, 18 \, i$$
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