Question

solve.
4x - 7 < x + 8
9 - x ≥ -3
\frac{x}{2} < -9
20 < 5x - 2
7x - 3x + 2 > x + 5x - 6
\frac{2x}{5}+6 ≤ 4
g(x) means “a number greater than x.” find a number to make each of these true.
g(4)=
g(-2.3)=
g(0)=
fill in the missing numbers in the table for each function.
f(x)=x - 5
\begin{tabular}{c|c}
x&f(x)\hline100&\hline10&\hline4&\hline1&\hline0&\hline-2&\hline-5&end{tabular}
g(x)=x^{2}+1
\begin{tabular}{c|c}
x&g(x)\hline20&\hline4&\hline1&\hline0&\hline-1&\hline-3&\hline-10&end{tabular}

Explanation:

Step1: Solve linear inequalities

For \(4x - 7

Step2: Solve linear - type inequalities with negative coefficients

For \(9 - x\geq - 3\), remember to reverse the sign when multiplying by - 1.

Step3: Solve fractional inequalities

For \(\frac{x}{2}<-9\), multiply by the denominator.

Step4: Solve inequalities with multiple steps

For \(20 < 5x-2\), first isolate the \(5x\) term.

Step5: Solve inequalities with like - term combination

For \(7x-3x + 2>x + 5x-6\), combine like - terms first.

Step6: Solve fractional inequalities with addition/subtraction

For \(\frac{2x}{5}+6\leq4\), isolate the fractional term first.

Step7: Find values for \(G(x)\)

Pick numbers greater than the given input for \(G(x)\).

Step8: Evaluate \(f(x)=x - 5\)

Substitute \(x\) values into the function \(f(x)\).

Step9: Evaluate \(g(x)=x^{2}+1\)

Substitute \(x\) values into the function \(g(x)\).

Answer:

1. Solve the inequalities:

• For \(4x - 7:
• Subtract \(x\) from both sides: \(4x-x-7
• Add 7 to both sides: \(3x-7 + 7<8 + 7\), so \(3x<15\).
• Divide both sides by 3: \(x < 5\).
• For \(9 - x\geq - 3\):
• Subtract 9 from both sides: \(9-9 - x\geq-3 - 9\), which gives \(-x\geq - 12\).
• Multiply both sides by - 1 (and reverse the inequality sign): \(x\leq12\).
• For \(\frac{x}{2}<-9\):
• Multiply both sides by 2: \(x<-18\).
• For \(20 < 5x-2\):
• Add 2 to both sides: \(20 + 2<5x-2 + 2\), so \(22<5x\).
• Divide both sides by 5: \(x>\frac{22}{5}=4.4\).
• For \(7x-3x + 2>x + 5x-6\):
• Combine like - terms on both sides. Left side: \(7x-3x+2 = 4x+2\), right side: \(x + 5x-6=6x-6\).
• So \(4x+2>6x-6\).
• Subtract \(4x\) from both sides: \(4x-4x+2>6x-4x-6\), which gives \(2>2x-6\).
• Add 6 to both sides: \(2 + 6>2x-6 + 6\), so \(8>2x\).
• Divide both sides by 2: \(4>x\) or \(x < 4\).
• For \(\frac{2x}{5}+6\leq4\):
• Subtract 6 from both sides: \(\frac{2x}{5}+6-6\leq4 - 6\), which gives \(\frac{2x}{5}\leq - 2\).
• Multiply both sides by 5: \(2x\leq-10\).
• Divide both sides by 2: \(x\leq - 5\).

1. For \(G(x)\):

• \(G(4)\): A number greater than 4, for example, 5.
• \(G(-2.3)\): A number greater than - 2.3, for example, - 2.
• \(G(0)\): A number greater than 0, for example, 1.

1. For \(f(x)=x - 5\):

• When \(x = 100\), \(f(100)=100-5 = 95\).
• When \(x = 10\), \(f(10)=10 - 5=5\).
• When \(x = 4\), \(f(4)=4 - 5=-1\).
• When \(x = 1\), \(f(1)=1 - 5=-4\).
• When \(x = 0\), \(f(0)=0 - 5=-5\).
• When \(x=-2\), \(f(-2)=-2 - 5=-7\).
• When \(x=-5\), \(f(-5)=-5 - 5=-10\).

1. For \(g(x)=x^{2}+1\):

• When \(x = 20\), \(g(20)=20^{2}+1=400 + 1=401\).
• When \(x = 4\), \(g(4)=4^{2}+1=16 + 1=17\).
• When \(x = 1\), \(g(1)=1^{2}+1=2\).
• When \(x = 0\), \(g(0)=0^{2}+1=1\).
• When \(x=-1\), \(g(-1)=(-1)^{2}+1=2\).
• When \(x=-3\), \(g(-3)=(-3)^{2}+1=9 + 1=10\).
• When \(x=-10\), \(g(-10)=(-10)^{2}+1=100 + 1=101\).