Solve the system of equations.

x - 3y + 2z = -11

2x - 4y + 3z = -15

3x - 5y - 4z = 5

x - 3y + 2z = -11

2x - 4y + 3z = -15

3x - 5y - 4z = 5

Dawson Downs
2022-07-27
Answered

Solve the system of equations.

x - 3y + 2z = -11

2x - 4y + 3z = -15

3x - 5y - 4z = 5

x - 3y + 2z = -11

2x - 4y + 3z = -15

3x - 5y - 4z = 5

You can still ask an expert for help

encoplemt5

Answered 2022-07-28
Author has **15** answers

x = 1

y = 2

z = -3

y = 2

z = -3

asked 2022-05-26

Find all solution in $\mathbb{R}$ for the following system of equations:

$\{\begin{array}{l}x+\frac{3x-y}{{x}^{2}+{y}^{2}}=3\\ y\u2013\frac{x+3y}{{x}^{2}+{y}^{2}}=0\end{array}$

$\{\begin{array}{l}x+\frac{3x-y}{{x}^{2}+{y}^{2}}=3\\ y\u2013\frac{x+3y}{{x}^{2}+{y}^{2}}=0\end{array}$

asked 2022-07-28

If the equation of a circle is ${x}^{2}+{y}^{2}={r}^{2}$ and the equation of a tangent line is y = mx + b,show that ${r}^{2}(1+{m}^{2})={b}^{2}$

asked 2022-11-10

Generalized way of solving this types of equations ${x}^{3}+{y}^{4}={z}^{5}$

asked 2022-04-28

Solve the system by substitution.

$x=4y-6$

$-4x+3y=-28$

asked 2022-09-05

Consider the coupled nonlinear system of equations given by

$${x}^{3}+{e}^{y}=s\text{}\text{}\text{}\text{}\text{}\text{}\text{}\mathrm{cos}x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

$${x}^{3}+{e}^{y}=s\text{}\text{}\text{}\text{}\text{}\text{}\text{}\mathrm{cos}x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

asked 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?

$2+{x}^{2}-{y}^{2}=0$

${x}^{2}-{y}^{2}-2=0$

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?

asked 2022-11-18

Solving system of equations: ${x}^{3}-3{y}^{2}x=-1$ and $3y{x}^{2}-{y}^{3}=1$