Respuesta :
x=fertilizer
3x+2=soil
1/2x=compost
Solve for x in order to solve each above.
x+(3x+2)+1/2x=29
4 1/2x+2=29
Subtract 2 from both sides
4 1/2x=27
9/2x=27
Multiply both sides by the inverse of 9/2 to get x by itself.
x=27*2/9
x=54/9
x=6=fertilizer
3x+2=3(6)+2=20=soil
1/2x=1/2(6)=3=compost
Check :
x+(3x+2)+1/2x=29
6+(3(6)+2)+1/2(6)=29
6+20+3=29
29=29
Hope this helps!! :)
3x+2=soil
1/2x=compost
Solve for x in order to solve each above.
x+(3x+2)+1/2x=29
4 1/2x+2=29
Subtract 2 from both sides
4 1/2x=27
9/2x=27
Multiply both sides by the inverse of 9/2 to get x by itself.
x=27*2/9
x=54/9
x=6=fertilizer
3x+2=3(6)+2=20=soil
1/2x=1/2(6)=3=compost
Check :
x+(3x+2)+1/2x=29
6+(3(6)+2)+1/2(6)=29
6+20+3=29
29=29
Hope this helps!! :)
Answer:17 kg
Step-by-step explanation:
let the weight of fertilizer be x
It is given that combined weight is 29 kg
Weight of soil was 2 kg more than three times the weight of fertilizer
i.e.
soil weight is 3x+2
Also weight of compost was half the weight of fertilizer
i.e. compost weight =[tex]\frac{1}{2}x[/tex]
And combined weight
x+3x+2+[tex]\frac{x}{2}[/tex]=29
x=6 kg
Difference in weight of soil and compost
[tex]3x+2-\frac{x}{2}=17 kg [/tex]