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High School
Algebra
Algebra I
Systems of equations

Recent questions in Systems of equations

Mara Cook 2022-06-16

Given $x + y$ and $x \cdot y$ , what is $x^{3} + y^{3}$ ?

Damon Stokes 2022-06-16

Are the following two systems of linear equations equivalent?
$- x_{1} + x_{2} + 4 x_{3} = 0$
$x_{1} + 3 x_{2} + 8 x_{3} = 0$
$\frac{1}{2} x_{1} + x_{2} + \frac{5}{2} x_{3} = 0$
and
$x_{1} - x_{3} = 0$
$x_{2} + 3 x_{3} = 0$
For the linear combination of the first equation of the second system, I got:
$x_{1} - x_{3} = x_{1} (- a + b + \frac{1}{2} c) + x_{2} (a + 3 b + c) + x_{3} (4 a + 8 b + \frac{5}{2} c)$
which results in this system of equations in order to solve for $a, b, c$ :
$(1) - a + b + \frac{1}{2} c = 1$
$(2) a + 3 b + c = 0$
$(3) 4 a + 8 b + \frac{5}{2} c = - 1$
Simplifying (1) and (2) gives you $4 b + \frac{3}{2} c = 1$ , and simplifying (2) and (3) gives you the same equation, which is why I figured there were infinite solutions. I used a different technique and different combination of equations, with the same result. (And the same thing happened with the second equation of the second system.)
Where did I go wrong?

Bailee Short 2022-06-16

Solution of 3 equations in 3 unknowns
Find the value of $c$ which makes it possible to solve:
$u + v + 2 w = 2,$
$2 u + 3 v - w = 5,$
$3 u + 4 v + w = c$

varitero5w 2022-06-16

Is the solution stable
$x^{'} = - x - 2 y + x^{2} y^{2}$ and $y^{'} = x - \frac{1}{2} y - \frac{1}{2} x^{3} y$
I have to check whether zero solution (solution for $x = y = 0$ ) is stable. Is it Lyapunov stable? Is it asymptotically stable?

sviraju6d 2022-06-16

Solving system if equations containing trigonometric functions
In trying to solve the following system of equation:
$20000 \times 9.81 + a \cos b = 0$
$a \sin b = 6.17 \times 20000$
Find $a$ and $b$ .
It gives me something containing "n2" in bold and I don't know why?
$a = - 231780.07 / (- 1)^{n 2}$
$b = (180 * (n 2 * p i - 0.56143476)) / p i$
How do I get the right answer?

Bailee Short 2022-06-15

Solving Two Equations
${\begin{aligned} 2 x = 13 - 5 y \\ 6 x + 15 y - 39 = 0 \end{aligned}$

Theresa Archer 2022-06-15

solve the following linear system
$x_{1} - 3 x_{2} + x_{3} = 1$
$2 x_{1} - x_{2} - 2 x_{3} = 2$
$x_{1} + 2 x_{2} - 3 x_{3} = - 1$
by using gauss-jordan elimination method. The augmented matrix of the linear system is
$(\begin{array}{cccc} 1 & - 3 & 1 & 1 \\ 2 & - 1 & - 2 & 2 \\ 1 & 2 & - 3 & - 1 \end{array}) .$
By a series of elementary row operations, we have
$(\begin{array}{cccc} 1 & - 3 & 1 & 1 \\ 0 & 5 & - 4 & 0 \\ 0 & 0 & 0 & - 2 \end{array}) .$
My question is, although the question asked us to solve the linear system using gauss-jordan elimination method, can we stop immediately and conclude that the linear system is inconsistent without further apply any elementary row operation to the matrix
$(\begin{array}{cccc} 1 & - 3 & 1 & 1 \\ 0 & 5 & - 4 & 0 \\ 0 & 0 & 0 & - 2 \end{array})$
until the matrix
$(\begin{array}{cccc} 1 & - 3 & 1 & 1 \\ 0 & 5 & - 4 & 0 \\ 0 & 0 & 0 & - 2 \end{array})$
is transformed into reduced-row echelon form?

aligass2004yi 2022-06-15

What is the solution of the following system?
$\begin{aligned} a \cdot e - b \cdot d & = α \\ a \cdot f - c \cdot d & = β \\ b \cdot f - c \cdot e & = γ \end{aligned}$
Where the unknowns $a, b, c, d, e, f$ are real numbers and $α, β, γ$ are fixed real numbers.

Armeninilu 2022-06-15

How to Find the all integer solutions for:
$x + y + z = 3$
$x^{3} + y^{3} + z^{3} = 3$

crossoverman9b 2022-06-15

How to divide solutions of system of ODE
$\begin{aligned} F_{1}^{'} = a F_{1} + b F_{2} \\ F_{2}^{'} = - b F_{1} + c F_{2} \end{aligned}$
for which I know the solutions of $F_{1}$ and $F_{2}$ as sums of exponential of eigenvalues. My question is can I find an ODE for $z = \frac{F_{1}}{F_{2}}$ . I want this so that I can find the following integral
$\int_{t_{0}}^{t} \frac{F_{1} (s)}{F_{2} (s)} d s$
explicitly.

kokoszzm 2022-06-14

Have the following system of equations:
$2 + x^{2} - y^{2} = 0$
$x^{2} - y^{2} - 2 = 0$
And if I substitute y by a function of x and vice versa I get:
$2 + x^{2} - x^{2} + 2 = 0$
$y^{2} - y^{2} - 4 = 0$
I therefore get:
$4 = 0, - 4 = 0$ Therefore I don't have any solutions for that system of equations
In theory, am I allowed to get to this conclusion?

Boilanubjaini8f 2022-06-14

How to solve these three equations?
If $α, β, γ$ are three numbers s.t.:
$α + β + γ = - 2 α^{2} + β^{2} + γ^{2} = 6 α^{3} + β^{3} + γ^{3} = - 5,$
then $α^{4} + β^{4} + γ^{4}$ is equal to ??

Leland Morrow 2022-06-14

Solving 3 simultaneous cubic equations
$i_{1}^{3} L_{1} + i_{1} K + V_{1} + (i_{2} + i_{3} + C) Z_{n} = 0$
$i_{2}^{3} L_{2} + i_{2} K + V_{2} + (i_{1} + i_{3} + C) Z_{n} = 0$
$i_{3}^{3} L_{3} + i_{3} K + V_{3} + (i_{1} + i_{2} + C) Z_{n} = 0$
where $L_{1}, L_{2}, L_{3}, K, V_{1}, V_{2}, V_{3}, C$ and $Z_{n}$ are all known constants.
What methods can I use to obtain the values of $i_{1}, i_{2}$ and $i_{3}$ ?

Jamiya Weber 2022-06-13

How to solve this simultaneous equations which involves exponents:
${\begin{cases} 2^{x} - 2^{y} = 1 \\ 4^{x} - 4^{y} = \frac{5}{3} \end{cases}$
Simultaneously. From the second equation I factorized to get
$(2^{x} + 2^{y}) (2^{x} - 2^{y}) = \frac{5}{3}$
and from the first equation, since $(2^{x} - 2^{y}) = 1$ , I deduced that
$(2^{x} + 2^{y}) \cdot 1 = \frac{5}{3} \to (2^{x} + 2^{y}) = \frac{5}{3}$
Now I am stuck on how to proceed.

Jase Howe 2022-06-13

How can I solve this system of non-linear equations?
${\begin{array}{lcr} a + c & = & 0 \\ b + a c + d & = & 6 \\ b c + a d & = & - 5 \\ b d & = & 6 \end{array}$
The book I'm working from provides the answer $a = - 1, b = 1, c = 1, d = 6$ , but I'm having a hard time getting there.
This is in order to reduce $q (x) = x^{4} + 6 x^{2} - 5 x + 6$ to a pair of irreducible quadratics $(x^{2} + a x + b) (x^{2} + c x + d)$ . We expand, gather like terms, and equate the coefficients of the resulting expression with those of the original, yielding the above system of equations.
With some substitutions, I managed to get
$b^{6} - 6 b^{5} - 6 b^{4} + 47 b^{3} - 36 b^{2} - 216 b + 216 = 0$
which I reduced to
$(b - 6) (b - 1) (b^{4} + b^{3} - 5 b^{2} + 6 b + 36) = 0$
Since the remaining factor $(b^{4} + b^{3} - 5 b^{2} + 6 b + 36)$ has no real zeros, it also reduces to a pair of irreducible quadratics.

Tristian Velazquez 2022-06-13

System of linear equations with parameters, using a matrix
Let there be the following system of linear equations:
$x + z + b w = a a x + y + a z + (a + a b) w = 1 + a^{2} b x + (a + b) z + (1 + b^{2}) w = 4 + a b x + b z + (a - a b + b^{2}) w = a + 1 + a b$
a,b parameters. The question is, for which a,b there is no solution to the system, for which there are infinite and for which there is one. I put it into a matrix and with some row operations I got to:
$\begin{array}{ccccc} 1 & 0 & 1 & b & a \\ 0 & 1 & 0 & a & 1 \\ 0 & 0 & a & 1 & 4 \\ 0 & 0 & 0 & a - 2 b & a + 1 \end{array}$
How do I continue from here? I'm quite confused. For instance I thought that saying that for all b, if a=2b then there is no solution.

Lydia Carey 2022-06-13

How to solve this system of equations
$4 x + 2 y + 4 z = 0$
$2 x + y + 2 z = 0$
$4 x + 2 y + 4 z = 0$

Theresa Archer 2022-06-13

Solving For Variables In Simultaneous Equations
1)
${\begin{cases} a^{2} + b^{2} = a \\ (a + d) b = b \\ b^{2} + d^{2} = d \end{cases}$
2)
${\begin{cases} a^{2} + b c = a \\ (a + d) b = b \\ (a + d) c = c \\ b c + d^{2} = d \end{cases}$

Boilanubjaini8f 2022-06-13

How to create a coefficient matrix from a system with sin and cos functions
$\sin (x_{1}) + 3 x_{2} + 4 x_{3} = 0$
$3 x_{1} + 5 x_{2} + \cos (x_{3}) = 0$
So it's clear to me, that the coefficients 3 and 4 are in first row in my coefficient matrix, but what about the sinus term? Same in the second equation with the cosinus term. Will they be in the column vector with my x's?

rigliztetbf 2022-06-12

Prove that $a = 2, b = 2, c = 2$ for a system of three equations.
Let us consider a system equations :
$\begin{matrix} (i) & a^{3} + b = 3 a + 4 \\ (ii) & 2 b^{3} + c = 6 b + 6 \\ (iii) & 3 c^{3} + a = 9 c + 8 \end{matrix}$
I have tried with the following steps :
$6 \times (i) + 3 \times (i i) + 2 \times$ (iii) gives:
$6 (a^{3} + b^{3} + c^{3}) + 6 b + 3 c + 2 a = 18 (a + b + c) + 58$ .
But I don't know how to solve in this way. By inspection method I have seen $a = b = c = 2$ .

In simple terms, systems of equations represent a special set of simultaneous equations where the equation system is used as a finite element. The trick here is to find common solutions, which is exactly what systems of equations solver must achieve. If this does not sound clear to you, take a look at some systems of equations answers below and see those with explanations. The solution will always come in three variables (namely, x, y, and z), which will represent your ordered triple. See systems of equations solutions for more examples of how it works in practice.

Algebra I

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