3X + 5 = 23
subtract 5 from each side
3X + 5 - 5 = 23 - 5
3X = 18
divide both sides integers by 3
(3/3)X = 18/3
X = 6
=======check by inserting 6 where X is in original equation
3(6) + 5 = 23
18 + 5 = 23
23 = 23
========checks
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z=pq
2z+6z+24=0 8z+24=0 8z= -24 z= - 3
why will the equations x+14=37 and x-14=37 have different solutions for x
Set up the equation and solve for z: 28 + z = 56 (next, subtract 28 from each side of the equaition to solve) z = 28
It is not possible to solve one equation in two unknowns (x and y). Two independent equations are required.