Question

Solve the system by substitution.5x - 2y + 3z =62x - 4y - 3z = - 23X + 6y - 8z = 1X =Y =Z =

Answer 1

X=1

Y=4

Z=3

Explanation

$\begin{gathered} 5x-2y+3z=6\Rightarrow equation(1) \\ 2x-4y-3z=-23\Rightarrow equation(1) \\ x+6y-8z=1\Rightarrow equation(1) \end{gathered}$

Step 1

b) isolate the x variable in equaiton (1) and equation(2)

so

$\begin{gathered} 5x-2y+3z=6\Rightarrow equation(1) \\ \text{add 2y in both sides} \\ 5x-2y+3z+2y=6+2y \\ 5x+3z=6+2y \\ \text{subtract 3z in both sides} \\ 5x+3z-3z=6+2y-3z \\ 5x=6+2y-3z \\ \text{divide both sides by 5} \\ \frac{5x}{5}=\frac{6+2y-3z}{5} \\ x=\frac{6+2y-3z}{5}\Rightarrow equation(1A) \end{gathered}$

and

$\begin{gathered} 2x-4y-3z=-23\Rightarrow equation(2) \\ 2x=-23+4y+3z \\ x=\frac{-23+4y+3z}{2} \end{gathered}$

b) now replace x value from equaton (1) into equation (3)

$\begin{gathered} x+6y-8z=1 \\ \frac{6+2y-3z}{5}+6y-8z=1 \\ \frac{6+2y-3z}{5}=1-6y+8z \\ 6+2y-3z=5(1-6y+8z) \\ 6+2y-3z=5-30y+40z \\ 6-5+2y+30y-3z-40z=0 \\ 1+32y-43z=0 \\ 32y-43z=-1\Rightarrow equation(4) \end{gathered}$

c) now replace x value from equaton (2) into equation (3)

$\begin{gathered} x+6y-8z=1 \\ \frac{-23+4y+3z}{2}+6y-8z=1 \\ \frac{-23+4y+3z}{2}=1-6y+8z \\ -23+4y+3z=2(1-6y+8z) \\ -23+4y+3z=2-12y+16z \\ -23+4y+3z-2+12y-16z=0 \\ -25+16y-13z=0 \\ 16y-13z=25\Rightarrow equation(5) \end{gathered}$

Step 2

now, use equation (4) and equation(5) to

$\begin{gathered} 32y-43z=-1\Rightarrow equation(4) \\ 16y-13z=25\Rightarrow equation(5) \end{gathered}$

a) isolate the y value from equation(1) and substitute into equation(2)

z=

$\begin{gathered} 32y-43z=-1\Rightarrow equation(4) \\ 32y=-1+43z \\ y=\frac{-1+43z}{32} \\ \text{replace in eq(5)} \\ 16y-13z=25\Rightarrow equation(5) \\ 16(\frac{-1+43z}{32})-13z=25 \\ (\frac{-1+43z}{2})-13z=25 \\ (\frac{-1+43z}{2})=25+13z \\ -1+43z=2(25+13z) \\ -1+43z=50+26z \\ 43z-26z=50+1 \\ 17z=51 \\ z=\frac{51}{17} \\ z=3 \end{gathered}$

so

Z= 3

b) replace the z value in equation (4) to find z

$\begin{gathered} 32y-43z=-1\Rightarrow equation(4) \\ 32y-43(3)=-1 \\ 32y=-1+129 \\ 32y=128 \\ y=4 \\ \end{gathered}$

Y=4

c) finally,l replace the Z and Y value in equation (1A)

$\begin{gathered} x=\frac{6+2y-3z}{5}\Rightarrow equation(1A) \\ x=\frac{6+2(4)-3(3)}{5} \\ x=\frac{6+8-9}{5}=\frac{5}{5}=1 \end{gathered}$

so

X=1

therefore, the answer is

X=1

Y=4

Z=3

I hope this helps you