In order to solve for the variable in the equation 2 (x + 3) + 5 x = 3 (2 x minus 1), Jaleesa begins by applying the distributiv
2 answers:
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Hello,
x,y,z∈ IN
==>
x∈{35,36,37}
y∈{47,48,49}
z∈{141,144,147}
We must go futher:
if x=35 then y=4*35/3=46.666 not good , not whole
if x=36 then y=4*36/3=48 ok
if x=37 then y=37*4/3=49.333.. not good
So there is only one solution (36,48,144) for (x,y,z)
No problem with z for z=3/1*4/3*x=4x ==>whole.
x,y,z∈ IN
x∈{35,36,37}
y∈{47,48,49}
z∈{141,144,147}
We must go futher:
if x=35 then y=4*35/3=46.666 not good , not whole
if x=36 then y=4*36/3=48 ok
if x=37 then y=37*4/3=49.333.. not good
So there is only one solution (36,48,144) for (x,y,z)
No problem with z for z=3/1*4/3*x=4x ==>whole.