Solve the system of equations 5x+5y-3z=1 15x+15y-9z=-1 2x-5y+5z=4 Select the correct choice below and fill in any answer boxes within your choice. A. There is one solution. The solution is (square .square .square ) (Type exact answers in simplified form.) B. There are infinitely many solutions. The solutions are (square ,square ,z) where z is any real number. (Type exact answers in simplified form.) C. There is no solution.

To solve the system of equations, we can use the method of elimination or substitution. Let's start by looking at the first two equations:<br /><br />$5x + 5y - 3z = 1$<br />$15x + 15y - 9z = -1$<br /><br />We can see that the second equation is just three times the first equation. This means that they are not independent of each other and represent the same plane. Therefore, we only need to consider the first equation and the third equation:<br /><br />$5x + 5y - 3z = 1$<br />$2x - 5y + 5z = 4$<br /><br />Now, let's solve these two equations using the elimination method. We can multiply the second equation by 5 to make the coefficients of y in both equations equal:<br /><br />$25x - 25y + 25z = 20$<br /><br />Now, let's add the two equations:<br /><br />$5x + 5y - 3z = 1$<br />$25x - 25y + 25z = 20$<br /><br />$30x + 22z = 21$<br /><br />Now, let's solve for x:<br /><br />$30x = 21 - 22z$<br />$x = \frac{21 - 22z}{30}$<br /><br />Now, substitute this value of x into the first equation to solve for y:<br /><br />$5(\frac{21 - 22z}{30}) + 5y - 3z = 1$<br /><br />$\frac{21 - 22z}{6} + 5y - 3z = 1$<br /><br />$\frac{21 - 22z + 30y - 18z}{6} = 1$<br /><br />$21 - 40z + 30y = 6$<br /><br />$30y = 40z - 15$<br /><br />$y = \frac{40z - 15}{30}$<br /><br />Now, we have expressions for x and y in terms of z. We can substitute these expressions back into either of the original equations to solve for z. Let's use the first equation:<br /><br />$5x + 5y - 3z = 1$<br /><br />$5(\frac{21 - 22z}{30}) + 5(\frac{40z - 15}{30}) - 3z = 1$<br /><br />$\frac{21 - 22z + 40z - 15}{6} - 3z = 1$<br /><br />$\frac{21 + 18z - 15}{6} - 3z = 1$<br /><br />$\frac{6 + 18z}{6} - 3z = 1$<br /><br />$1 + 3z - 3z = 1$<br /><br />$1 = 1$<br /><br />This equation is always true, which means that there are infinitely many solutions for z. Therefore, the correct choice is:<br /><br />B. There are infinitely many solutions. The solutions are $(\frac{21 - 22z}{30}, \frac{40z - 15}{30}, z)$ where z is any real number.