Question

determinati numarul real m pentru care ecuatia:a) 2x+m=4x+4 are solutia -2
b)ecuatia m(x+2) +3(x-1)=mx-3 ,are solutia 0
c) 2x-m(x+3) =7mx+12 are solutia 5
d) -3x +4(mx-1)=6x+2-7(m+2) are solutia 3
3(m-1)(x+2) -2(2m-1)(2x+1) =7x+3mx-5 are solutia -3

Answer 1

a) ecuatia 2x+m=4x+4 are solutia -2
2×(-2)+m=4×(-2)+4
-4+m=-8+4
-4+m=-4
m=0

b)ecuatia m(x+2) +3(x-1)=mx-3 ,are solutia 0
m(0+2) +3(0-1)=m×0-3
2m -3=0-3
2m =0
m=0

c) ecuatia 2x-m(x+3) =7mx+12 are solutia 5
2×5-m(5+3) =7m×5+12
10-8m =35m+12
43m=-2
m=-2/43

d)ecuatia  -3x +4(mx-1)=6x+2-7(m+2) are solutia 3
-3×3 +4(m×3-1)=6×3+2-7(m+2)
-9+12m-4=18+2-7m-14
12m-13=6-7m
19m=19
m=1

e)ecuatia 3(m-1)(x+2) -2(2m-1)(2x+1) =7x+3mx-5 are solutia -3 3(m-1)(-3+2) -2(2m-1)[2×(-3)+1] =7×(-3)+3m×(-3)-5
-3(m-1) +10(2m-1) =-21-9m-5
-3m+3 +20m-10 =-21-9m-5
17m-7=-9m-26
26m=-19
m=-19/26