How do you solve the following system of equations 3x +2y -z = 17.8, x - 3y + 2z = 7.9, 2x + y - 3z = 3.9?
[math]\begin{align} 3x+2y-z &=& 17.8 \tag{a} \\ x-3y+2z &=& 7.9 \tag{b} \\ 2x+y-3z &=& 3.9 \tag{c} \end{align}[/math][math]\text{d: a-3b}=11y-7y=-5.9[/math][math]\text{e: 2a-3c}=y+7z=23.9[/math][math]\text{d+e}=12y=18 \implies y=\frac{18}{12}=\frac{3}{2}=1.5[/math]using this result in [math]e[/math]:[math]1.5+7z=23.9 \implies 7z=22.4 \implies z=\frac{224}{70}=\frac{16}{5}=3.2[/math]And finally using all of these in [math]b[/math]:[math]x=7.9+3 \cdot 1.5-2 \cdot 3.2=6[/math]But honestly I just used Websites to do it:Wolfram AlphaMatrix calculator
Solve the following system of equations. Please show your work to receive full credit.?
x- y = 13 (Solve for x) x = y + 13 Now use substitution in 2nd equation to find y 2(13 + y) + y = 2 (substitution) 26 + 2y + y = 2 (distribution) 26 + 3y = 2 (combine like terms) 3y = -24 (subtract 26 from both sides) y = -8 (divide both sides by 3 to solve for y) Now put y back into the original equation to find x x - (-8) = 13 (substitution) x + 8 = 13 (negative cancels and makes it addition) x = 5 (subtract 8 from both sides to solve for x) x = 5 and y = -8