Solve the following system. ) x+3y=1 -2y-z=1 4x-3z=13 x= square y.= square z= square

To solve the system of equations, we can use the method of substitution or elimination. Let's use the elimination method.<br /><br />First, let's eliminate $z$ from the first and third equations. Multiply the second equation by 3 to make the coefficients of $y$ in the second and third equations equal:<br /><br />$\{ \begin{matrix} x+3y=1\\ -6y-3z=3\\ 4x-3z=13\end{matrix} $<br /><br />Now, add the second and third equations to eliminate $z$:<br /><br />$(x+3y)+(4x-3z)=1+13$<br /><br />Simplify:<br /><br />$5x+3y=14$<br /><br />Now, we have a new system of equations:<br /><br />$\{ \begin{matrix} x+3y=1\\ 5x+3y=14\end{matrix} $<br /><br />Subtract the first equation from the second equation to eliminate $y$:<br /><br />$(5x+3y)-(x+3y)=14-1$<br /><br />Simplify:<br /><br />$4x=13$<br /><br />Divide both sides by 4:<br /><br />$x=\frac{13}{4}$<br /><br />Now that we have the value of $x$, we can substitute it back into the first equation to solve for $y$:<br /><br />$\frac{13}{4}+3y=1$<br /><br />Subtract $\frac{13}{4}$ from both sides:<br /><br />$3y=1-\frac{13}{4}$<br /><br />Simplify:<br /><br />$3y=\frac{4}{4}-\frac{<br /><br />$3y=-\frac{9}{4}$<br /><br />Divide both sides by 3:<br /><br />$y=-\frac{3}{4}$<br /><br />Finally, we can substitute the values of $x$ and $y$ into the second equation to solve for $z$:<br /><br />$-2(-\frac{3}{4})-z=1$<br /><br />Simplify:<br /><br />$\frac{3}{2}-z=1$<br /><br />Subtract $\frac{3}{2}$ from both sides:<br /><br />$-z=1-\frac{3}{2}$<br /><br />Simplify:<br /><br />$-z=\frac{2}{2}-\frac{3}{2}$<br /><br />$-z=-\frac{1}{2}$<br /><br />Multiply both sides by -1:<br /><br />$z=\frac{1}{2}$<br /><br />Therefore, the solution to the system of equations is:<br /><br />$x=\frac{13}{4}$<br /><br />$y=-\frac{3}{4}$<br /><br />$z=\frac{1}{2}$