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GuDViN [60]
3 years ago
15

How do i solve 1/4 (x+3)-4x<3 (1+x) ... its suppose to be a greater than or equal to sign

Mathematics
2 answers:
pav-90 [236]3 years ago
8 0
      ¹/₄(x + 3) - 4x ≤ 3(1 + x)
¹/₄(x) + ¹/₄(3) - 4x ≤ 3(1) + 3(x)
      ¹/₄x + ³/₄ - 4x ≤ 3 + 3x
      ¹/₄x - 4x + ³/₄ ≤ 3 + 3x
          -3³/₄x + ³/₄ ≤ 3 + 3x
        <u>+ 3³/₄x            + 3³/₄x
</u>                      ³/₄ ≤ 3 + 6³/₄x
                     <u>- 3   - 3            
</u><u />                   <u>-2¹/₄</u> ≤ <u>6³/₄x</u>
                    6³/₄     6³/₄
                      ⁻¹/₃ ≤ x
                         x ≥ ⁻¹/₃
sdas [7]3 years ago
7 0
First, multiply 1/4 by x+3 ⇒(1/4x+3/4). Next, multiply 3 by 1+x ⇒(3+3x). The equation should now look like this: 1/4x+3/4-4x≤3+3x. The next thing you have to do is get all of the x's on one side of the symbol. Subtract 1/4x from both sides, and the equation then looks like this: 3/4-4x≤3+2 3/4x. Add the -4x to both sides, and the equation should look like this: 3/4≤3+6 3/4x. Next, you have to move the 3 to the left side. To do that, subtract it from both sides. The equation then look like this: -2 1/4≤6 3/4x. Then, to get the x alone, you need to divide the coefficient (6 3/4) by -2 1/4, which equals -1/3 or -0.3 repeating. So, -1/3≤x
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