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Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Select two options.

x < 5
–6x – 5 < 10 – x
–6x + 15 < 10 – 5x
A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right.
A number line from negative 3 to 3 in increments of 1. An open circle is at negative 5 and a bold line starts at negative 5 and is pointing to the left.

Respuesta :

semsee45

Answer:

–6x + 15 < 10 – 5x

A number line from negative 3 to 7 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right.

Step-by-step explanation:

Given inequality

[tex]-3(2x-5) < 5(2-x)[/tex]

Solving the inequality

Expand the brackets:

[tex]\implies -6x+15 < 10-5x[/tex]

Add 6x to both sides:

[tex]\implies 15 < 10+x[/tex]

Subtract 10 from both sides:

[tex]\implies 5 < x[/tex]

[tex]\implies x > 5[/tex]

Therefore, x is bigger than 5.

Graphing the inequality

When graphing inequalities on a number line:

  • < or > = open circle
  • ≤ or ≥ = closed circle
  • < or ≤ = shade to the left
  • > or ≥ = shade to the right

To graph the given inequality on a number line:

  • place an open circle at 5
  • draw a line starting at 5 and pointing to the right

Conclusion

Therefore, the correct representations of the given inequality are:

  • –6x + 15 < 10 – 5x
  • A number line from negative 3 to 7 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right.
Ver imagen semsee45

[tex]\\ \rm\leadsto -3(2x-5)<5(2-x)[/tex]

  • a(b+c)=ab+ac

[tex]\\ \rm\leadsto -6x+15<10-5x[/tex]

Option C

[tex]\\ \rm\leadsto -x<-5[/tex]

[tex]\\ \rm\leadsto x>5[/tex]

Rest are wrong

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