Need help solving 20< 5x < 35?

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Please show work!!!

2007-03-29 08:30:20 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

20 < 5x < 35
Divide each side by 5
20/5 < 5x/5 < 35/5
4 < x < 7

2007-03-29 08:33:38 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋

Divide throughout by 5:
20/5 < 5x/5 < 35/5
4 < 1x < 7
4 < x < 7

2007-03-29 15:35:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋

2007-03-29 15:39:34 · answer #3 · answered by bigfanbudselect 1 · 0⤊ 0⤋

Just make inequalities = and solve. 20=5x x=4 5x=35 x=7
answer 4

2007-03-29 15:42:16 · answer #4 · answered by dwinbaycity 5 · 0⤊ 0⤋

Just divide by 5 getting :
4

2007-03-29 15:36:28 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋

Do each bit separately
20<5x
so divide both sides by 5 gives
4
5x<35
so divide both sides by 5 gives
x<7

so finally 4

2007-03-29 15:37:36 · answer #6 · answered by fuggle01 3 · 0⤊ 0⤋

20 < 5x < 35

20 / 5 < 5x / 5 < 35 / 5

4 < x < 7

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2007-03-29 16:12:24 · answer #7 · answered by SAMUEL D 7 · 0⤊ 0⤋

"x" can be 5 or 6 if there are no constraints given ... if there are any constraints .. the answer may vary ...

thanks

2007-03-29 15:35:56 · answer #8 · answered by Interpol 1 · 0⤊ 1⤋