7x+5y-3z =16, 3x-5y+2z=-8, 5x+3y-7z=0. Solve for x,y and z.

There are three methods to solve these types of silmultaneous equation questions, substitutution, elimination and matrices. In this example we use substitution and elimination.Elimination example.Eq1). 7x+5y-3z =16Eq2). 3x-5y+2z=-8Eq3). 5x+3y-7z=0
Eq4). Eq1 + Eq2 (chosen on inspection because +5y and -5y will cancel each other out)Eq4). 10x-z=8
Now we need another equation with just two variables.
To remove the x values when adding equations 2 and 3 we need to multiple each equation. Eq2. contains -5y, Eq3. contains +3y. Therefore multiplying Eq2. by 3 and Eq3. by 5. Will give us +15y and -15y respectively.
Eq5). 3(Eq2) + 5(Eq3)Eq5). 3[3x-5y+2z=-8] + 5[5x+3y-7z=0]Eq5). [9x-15y+6z=-24] + [25x+15y-35z=0]Eq5). 34x-29z=-24
Now we have two equations with only two variables.Eq4). 10x-z=8Eq5). 34x-29z=-24
To solve for x through substitution.Eq4). Can be rearranged to... z = 10x-8Which can be substituted into Eq5). making Eq6).Eq6). 34x - 29(10x-8) = -24Eq6). 34x - 290x - - 232 = -24Eq6). -256x + 232 = -24Eq6). -256x = -256Therefore dividing by -256, x = 1.
Substituting back into the rearranged Eq4). z = 10x-8 where x=1.z = 10(1) -8z = 10-8z = 2
To work out the y variable we will go back to the original Eq1) and solve with the values for x and z.Eq1). 7x+5y-3z =16, where x=1 and z=27(1) + 5y - 3(2) = 16,7 + 5y - 6 = 16,1 + 5y = 16,5y = 15,y=3
Final Answer, x=1, y=3, z=2