Systems of 3 equations by using a combination of both substitution and elimination

X-2y+z=3
2x+y-2z=31
-x+2y+3z=-23
multiply the first equation by -2 and add the results to the second equation:
x-2y+z=3
+5y-4z=25
-x+2y+3z=-23
multiply the first equation by 1 and the results to the third equation:
 x-2y+z=3
5y-4z=25
4z=-20
solve for z:
4z=-20
z=-5
solve for y:
5y-4z=25,z=-5
5y-(4*-5)=25
5y+20=25
5y+20-20=25-20
5y=5
y=1
solve for x by substituted y and z into the first equation:
x-2y+z=3, y=1, z=-5
x-(2*1)-5=3
x-2-5=3
x-7=3
x-7+7=3+7
x=10
the answer are x=10, y=1, z=-5

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.

Bi’m pretty sure hope it works