Answer:
Part 1) [tex]x\geq10[/tex]
Part 2) [tex]m\leq -9[/tex]
Part 3) [tex]p\geq 5[/tex]
Part 4) [tex]x<-10[/tex]
Part 5) [tex]b<-10[/tex]
Part 6) [tex]n<5[/tex]
Part 7) [tex]n <6[/tex]
Part 8) [tex]r\leq 4[/tex]
Part 9) [tex]x\geq 7[/tex]
Part 10) [tex]p\leq 0[/tex]
Part 11) [tex]x<1[/tex]
Part 12) [tex]a > 24[/tex]
Step-by-step explanation:
Part 1) [tex]2x+4\geq24[/tex]
Subtract 4 both sides
[tex]2x\geq24-4[/tex]
[tex]2x\geq20[/tex]
Divide by 2 both sides
[tex]x\geq10[/tex]
the solution is the interval ------> [10,∞)
The solution is the shaded area to the right of the solid line at number 10 (closed circle).
see the attached figure
Part 2) [tex]\frac{m}{3}-3\leq -6[/tex]
Adds 3 both sides
[tex]\frac{m}{3}\leq -6+3[/tex]
[tex]\frac{m}{3}\leq -3[/tex]
Multiply by 3 both sides
[tex]m\leq -9[/tex]
the solution is the interval ------> (-∞,-9]
The solution is the shaded area to the left of the solid line at number -9 (closed circle).
see the attached figure
Part 3) [tex]-3(p+1)\leq -18[/tex]
applying the distributive property left side
[tex]-3p-3\leq -18[/tex]
adds 3 both sides
[tex]-3p\leq -18+3[/tex]
[tex]-3p\leq -15[/tex]
Multiply by -1 both sides
[tex]3p\geq 15[/tex]
Divide by 3 both sides
[tex]p\geq 5[/tex]
the solution is the interval ------> [5,∞)
The solution is the shaded area to the right of the solid line at number 5 (closed circle).
see the attached figure
Part 4) [tex]-4(-4+x)>56[/tex]
applying the distributive property left side
[tex]16-4x>56[/tex]
Subtract 16 both sides
[tex]-4x>56-16[/tex]
[tex]-4x>40[/tex]
Multiply by -1 both sides
[tex]4x<-40[/tex]
Divide by 4 both sides
[tex]x<-10[/tex]
the solution is the interval ------> (-∞,-10)
The solution is the shaded area to the left of the dashed line at number -10 (open circle).
see the attached figure
Part 5) [tex]-b-2>8[/tex]
adds 2 both sides
[tex]-b>8+2[/tex]
[tex]-b>10[/tex]
Multiply by -1 both sides
[tex]b<-10[/tex]
the solution is the interval ------> (-∞,-10)
The solution is the shaded area to the left of the dashed line at number -10 (open circle).
Part 6) [tex]-4(3+n)>-32[/tex]
applying the distributive property left side
[tex]-12-4n>-32[/tex]
adds 12 both sides
[tex]-4n>-32+12[/tex]
[tex]-4n>-20[/tex]
multiply by -1 both sides
[tex]4n<20[/tex]
divide by 4 both sides
[tex]n<5[/tex]
the solution is the interval ------> (-∞,5)
The solution is the shaded area to the left of the dashed line at number 5 (open circle).
see the attached figure
Part 7) [tex]4+\frac{n}{3} <6[/tex]
Subtract 4 both sides
[tex]\frac{n}{3} <6-4[/tex]
[tex]\frac{n}{3} <2[/tex]
Multiply by 3 both sides
[tex]n <6[/tex]
the solution is the interval ------> (-∞,6)
The solution is the shaded area to the left of the dashed line at number 6 (open circle).
see the attached figure
Part 8) [tex]-3(r-4)\geq 0[/tex]
applying the distributive property left side
[tex]-3r+12\geq 0[/tex]
subtract 12 both sides
[tex]-3r\geq -12[/tex]
divide by -1 both sides
[tex]3r\leq 12[/tex]
divide by 3 both sides
[tex]r\leq 4[/tex]
the solution is the interval ------> (-∞,4]
The solution is the shaded area to the left of the solid line at number 4 (closed circle).
see the attached figure
Part 9) [tex]-7x-7\leq -56[/tex]
Adds 7 both sides
[tex]-7x\leq -56+7[/tex]
[tex]-7x\leq -49[/tex]
Multiply by -1 both sides
[tex]7x\geq 49[/tex]
Divide by 7 both sides
[tex]x\geq 7[/tex]
the solution is the interval ------> [7,∞)
The solution is the shaded area to the right of the solid line at number 7 (closed circle).
see the attached figure
Part 10) [tex]-3(p-7)\geq 21[/tex]
applying the distributive property left side
[tex]-3p+21\geq 21[/tex]
subtract 21 both sides
[tex]-3p\geq 21-21[/tex]
[tex]-3p\geq 0[/tex]
Multiply by -1 both sides
[tex]3p\leq 0[/tex]
[tex]p\leq 0[/tex]
the solution is the interval ------> (-∞,0]
The solution is the shaded area to the left of the solid line at number 0 (closed circle).
see the attached figure
Part 11) [tex]-11x-4> -15[/tex]
Adds 4 both sides
[tex]-11x> -15+4[/tex]
[tex]-11x> -11[/tex]
Multiply by -1 both sides
[tex]11x<11[/tex]
Divide by 11 both sides
[tex]x<1[/tex]
the solution is the interval ------> (-∞,1)
The solution is the shaded area to the left of the dashed line at number 1 (open circle).
see the attached figure
Part 12) [tex]\frac{-9+a}{15}>1[/tex]
Multiply by 15 both sides
[tex]-9+a > 15[/tex]
Adds 9 both sides
[tex]a > 15+9[/tex]
[tex]a > 24[/tex]
the solution is the interval ------> (24,∞)
The solution is the shaded area to the right of the dashed line at number 24 (open circle).
see the attached figure
