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${:[" j) "-(3)/(3+4y)=(1)/(x-z)],[(3)/(x+y-1)=-z^(-1)],[(6z)/(5y+8)=-1//:5y+8]:}$

$\begin{array}{l}\text { j) }-\frac{3}{3+4 y}=\frac{1}{x-z} \\ \frac{3}{x+y-1}=-z^{-1} \\ \frac{6 z}{5 y+8}=-1 /: 5 y+8\end{array}$

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Calculus

Find higher derivatives of rational and radical functions

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\begin{array}{l}\text { j) }-\frac{3}{3+4 y}=\frac{1}{x-z} \\ \frac{3}{x+y-1}=-z^{-1} \\ \frac{6 z}{5 y+8}=-1 /: 5 y+8\end{array}

Simplify First Equation: To solve the system of equations, we will first simplify each equation and then attempt to solve for the variables x, y, and z. $\newline$ Let's start with the first equation: $\newline$ $-(3)/(3+4y)=(1)/(x-z)$ $\newline$ We can cross-multiply to get rid of the fractions: $\newline$ $-3(x-z) = 1(3+4y)$
Distribute $-3$ and $1$ : Now, let's distribute the $-3$ and the $1$ on both sides of the equation: $\newline$ $-3x + 3z = 3 + 4y$
Simplify Second Equation: Next, we move on to the second equation: $\newline$ $(3)/(x+y-1)=-z^{-1}$ $\newline$ Since $-z^{-1}$ is the same as $-1/z$ , we can cross-multiply to eliminate the fractions: $\newline$ $3z = -1(x+y-1)$
Distribute $-1$ : Now, distribute the $-1$ on the right side of the equation: $\newline$ $3z = -x - y + 1$
Simplify Third Equation: Let's move to the third equation: $\newline$ $(6z)/(5y+8)=-1$ $\newline$ We can multiply both sides by $(5y+8)$ to get rid of the fraction: $\newline$ $6z = -1(5y+8)$
Distribute $-1$ : Now, distribute the $-1$ on the right side of the equation: $\newline$ $6z = -5y - 8$
Check Consistency: We now have three simplified equations: $\newline$ $1$ . $-3x + 3z = 3 + 4y$ $\newline$ $2$ . $3z = -x - y + 1$ $\newline$ $3$ . $6z = -5y - 8$ $\newline$ We can use these equations to solve for x, y, and z. However, we notice that the equations are not linearly independent. For instance, multiplying the second equation by $2$ gives us the third equation. This means that the system of equations may have infinitely many solutions or no solution at all, depending on the consistency of the equations.
Check Consistency: We now have three simplified equations: $\newline$ $1$ . $-3x + 3z = 3 + 4y$ $\newline$ $2$ . $3z = -x - y + 1$ $\newline$ $3$ . $6z = -5y - 8$ $\newline$ We can use these equations to solve for x, y, and z. However, we notice that the equations are not linearly independent. For instance, multiplying the second equation by $2$ gives us the third equation. This means that the system of equations may have infinitely many solutions or no solution at all, depending on the consistency of the equations.To determine the number of solutions, we need to check if the equations are consistent. If they are, the system has infinitely many solutions; if not, it has no solution. We can do this by comparing the ratios of the coefficients of z in the second and third equations. $\newline$ From the second equation, the coefficient of z is $3$ , and from the third equation, the coefficient of z is $6$ . The ratio is $3$ : $6$ or $1$ : $2$ . Now, let's compare the ratios of the constants in the same equations. $\newline$ From the second equation, the constant is $1$ , and from the third equation, the constant is $-8$ . The ratio is $1$ : $-8$ , which is not the same as $1$ : $2$ . This inconsistency indicates that the system of equations has no solution.