Compound inequalities combine two inequalities using the words “and” or “or.” An “and” inequality requires both conditions to be true, so its solution is the overlap of the two solution sets, while an “or” inequality requires at least one condition to be true. To solve compound inequalities, isolate the variable, remembering to reverse the inequality signs when multiplying or dividing by a negative number. Solutions can be written as inequalities and graphed on a number line using open or closed circles.
Notes
Questions
Solve and graph
\(\textbf{1)}\) \(-4\lt4x+6\le18\)The answer is \(-\frac{5}{2}\lt x\le3\)
\(\,\,\,\,\,\,-4\lt4x+6\le18\)
Subtract 6 from all three parts.
\(\,\,\,\,\,\,-10\lt4x\le12\)
Divide all three parts by 4.
\(\,\,\,\,\,\,-\frac{10}{4}\lt x\le\frac{12}{4}\)
\(\,\,\,\,\,\,-\frac{5}{2}\lt x\le3\)
Graph: Open circle at \(-\frac{5}{2}\) and closed circle at 3. Shade between them.
\(\,\,\,\,\,\,-4\lt4x+6\le18\)
Subtract 6 from all three parts.
\(\,\,\,\,\,\,-10\lt4x\le12\)
Divide all three parts by 4.
\(\,\,\,\,\,\,-\frac{10}{4}\lt x\le\frac{12}{4}\)
\(\,\,\,\,\,\,-\frac{5}{2}\lt x\le3\)
Graph: Open circle at \(-\frac{5}{2}\) and closed circle at 3. Shade between them.
\(\textbf{2)}\) \(5x-4\lt34\) and \(-5x-8\le-8\)The answer is \(0\le x\lt\frac{38}{5}\)
Solve the first inequality.
\(\,\,\,\,\,\,5x-4\lt34\)
\(\,\,\,\,\,\,5x\lt38\)
\(\,\,\,\,\,\,x\lt\frac{38}{5}\)
Solve the second inequality.
\(\,\,\,\,\,\,-5x-8\le-8\)
\(\,\,\,\,\,\,-5x\le0\)
Divide by -5 and reverse the inequality sign.
\(\,\,\,\,\,\,x\ge0\)
Since this is an “and” inequality, find the overlap.
\(\,\,\,\,\,\,0\le x\lt\frac{38}{5}\)
Graph: Closed circle at 0 and open circle at \(\frac{38}{5}\). Shade between them.
Solve the first inequality.
\(\,\,\,\,\,\,5x-4\lt34\)
\(\,\,\,\,\,\,5x\lt38\)
\(\,\,\,\,\,\,x\lt\frac{38}{5}\)
Solve the second inequality.
\(\,\,\,\,\,\,-5x-8\le-8\)
\(\,\,\,\,\,\,-5x\le0\)
Divide by -5 and reverse the inequality sign.
\(\,\,\,\,\,\,x\ge0\)
Since this is an “and” inequality, find the overlap.
\(\,\,\,\,\,\,0\le x\lt\frac{38}{5}\)
Graph: Closed circle at 0 and open circle at \(\frac{38}{5}\). Shade between them.
\(\textbf{3)}\) \(3\lt6-2x\le18\)The answer is \(-6\le x\lt\frac{3}{2}\)
\(\,\,\,\,\,\,3\lt6-2x\le18\)
Subtract 6 from all three parts.
\(\,\,\,\,\,\,-3\lt-2x\le12\)
Divide all three parts by -2 and reverse both inequality signs.
\(\,\,\,\,\,\,\frac{3}{2}\gt x\ge-6\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-6\le x\lt\frac{3}{2}\)
Graph: Closed circle at -6 and open circle at \(\frac{3}{2}\). Shade between them.
\(\,\,\,\,\,\,3\lt6-2x\le18\)
Subtract 6 from all three parts.
\(\,\,\,\,\,\,-3\lt-2x\le12\)
Divide all three parts by -2 and reverse both inequality signs.
\(\,\,\,\,\,\,\frac{3}{2}\gt x\ge-6\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-6\le x\lt\frac{3}{2}\)
Graph: Closed circle at -6 and open circle at \(\frac{3}{2}\). Shade between them.
\(\textbf{4)}\) \(4x-5\lt35\) and \(4-x\le-10\)The answer is No Solution
Solve the first inequality.
\(\,\,\,\,\,\,4x-5\lt35\)
\(\,\,\,\,\,\,4x\lt40\)
\(\,\,\,\,\,\,x\lt10\)
Solve the second inequality.
\(\,\,\,\,\,\,4-x\le-10\)
\(\,\,\,\,\,\,-x\le-14\)
Divide by -1 and reverse the inequality sign.
\(\,\,\,\,\,\,x\ge14\)
Combine the solutions using “and.”
\(\,\,\,\,\,\,x\lt10\) and \(x\ge14\)
No number can satisfy both inequalities.
The answer is No Solution.
Graph: The two solution sets do not overlap, so no points are shaded in the combined solution.
Solve the first inequality.
\(\,\,\,\,\,\,4x-5\lt35\)
\(\,\,\,\,\,\,4x\lt40\)
\(\,\,\,\,\,\,x\lt10\)
Solve the second inequality.
\(\,\,\,\,\,\,4-x\le-10\)
\(\,\,\,\,\,\,-x\le-14\)
Divide by -1 and reverse the inequality sign.
\(\,\,\,\,\,\,x\ge14\)
Combine the solutions using “and.”
\(\,\,\,\,\,\,x\lt10\) and \(x\ge14\)
No number can satisfy both inequalities.
The answer is No Solution.
Graph: The two solution sets do not overlap, so no points are shaded in the combined solution.
\(\textbf{5)}\) \(x+3\gt8\) or \(2x\lt16\)The answer is All Real Numbers
Solve the first inequality.
\(\,\,\,\,\,\,x+3\gt8\)
\(\,\,\,\,\,\,x\gt5\)
Solve the second inequality.
\(\,\,\,\,\,\,2x\lt16\)
\(\,\,\,\,\,\,x\lt8\)
Combine the solutions using “or.”
\(\,\,\,\,\,\,x\gt5\) or \(x\lt8\)
Every real number satisfies at least one of these inequalities.
The answer is All Real Numbers.
Graph: Shade the entire number line.
Solve the first inequality.
\(\,\,\,\,\,\,x+3\gt8\)
\(\,\,\,\,\,\,x\gt5\)
Solve the second inequality.
\(\,\,\,\,\,\,2x\lt16\)
\(\,\,\,\,\,\,x\lt8\)
Combine the solutions using “or.”
\(\,\,\,\,\,\,x\gt5\) or \(x\lt8\)
Every real number satisfies at least one of these inequalities.
The answer is All Real Numbers.
Graph: Shade the entire number line.
\(\textbf{6)}\) \(2\lt x+5\le11\)The answer is \(-3\lt x\le6\)
\(\,\,\,\,\,\,2\lt x+5\le11\)
Subtract 5 from all three parts.
\(\,\,\,\,\,\,2-5\lt x\le11-5\)
\(\,\,\,\,\,\,-3\lt x\le6\)
Graph: Open circle at -3 and closed circle at 6. Shade between them.
\(\,\,\,\,\,\,2\lt x+5\le11\)
Subtract 5 from all three parts.
\(\,\,\,\,\,\,2-5\lt x\le11-5\)
\(\,\,\,\,\,\,-3\lt x\le6\)
Graph: Open circle at -3 and closed circle at 6. Shade between them.
\(\textbf{7)}\) \(-7\le3x+2\lt14\)The answer is \(-3\le x\lt4\)
\(\,\,\,\,\,\,-7\le3x+2\lt14\)
Subtract 2 from all three parts.
\(\,\,\,\,\,\,-9\le3x\lt12\)
Divide all three parts by 3.
\(\,\,\,\,\,\,-3\le x\lt4\)
Graph: Closed circle at -3 and open circle at 4. Shade between them.
\(\,\,\,\,\,\,-7\le3x+2\lt14\)
Subtract 2 from all three parts.
\(\,\,\,\,\,\,-9\le3x\lt12\)
Divide all three parts by 3.
\(\,\,\,\,\,\,-3\le x\lt4\)
Graph: Closed circle at -3 and open circle at 4. Shade between them.
\(\textbf{8)}\) \(-10\lt-2x+4\le8\)The answer is \(-2\le x\lt7\)
\(\,\,\,\,\,\,-10\lt-2x+4\le8\)
Subtract 4 from all three parts.
\(\,\,\,\,\,\,-14\lt-2x\le4\)
Divide all three parts by -2 and reverse both inequality signs.
\(\,\,\,\,\,\,7\gt x\ge-2\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-2\le x\lt7\)
Graph: Closed circle at -2 and open circle at 7. Shade between them.
\(\,\,\,\,\,\,-10\lt-2x+4\le8\)
Subtract 4 from all three parts.
\(\,\,\,\,\,\,-14\lt-2x\le4\)
Divide all three parts by -2 and reverse both inequality signs.
\(\,\,\,\,\,\,7\gt x\ge-2\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-2\le x\lt7\)
Graph: Closed circle at -2 and open circle at 7. Shade between them.
\(\textbf{9)}\) \(1\le\frac{x-2}{3}\lt5\)The answer is \(5\le x\lt17\)
\(\,\,\,\,\,\,1\le\frac{x-2}{3}\lt5\)
Multiply all three parts by 3.
\(\,\,\,\,\,\,3\le x-2\lt15\)
Add 2 to all three parts.
\(\,\,\,\,\,\,5\le x\lt17\)
Graph: Closed circle at 5 and open circle at 17. Shade between them.
\(\,\,\,\,\,\,1\le\frac{x-2}{3}\lt5\)
Multiply all three parts by 3.
\(\,\,\,\,\,\,3\le x-2\lt15\)
Add 2 to all three parts.
\(\,\,\,\,\,\,5\le x\lt17\)
Graph: Closed circle at 5 and open circle at 17. Shade between them.
\(\textbf{10)}\) \(-2\le\frac{x+1}{-3}\lt4\)The answer is \(-13\lt x\le5\)
\(\,\,\,\,\,\,-2\le\frac{x+1}{-3}\lt4\)
Multiply all three parts by -3 and reverse both inequality signs.
\(\,\,\,\,\,\,6\ge x+1\gt-12\)
Subtract 1 from all three parts.
\(\,\,\,\,\,\,5\ge x\gt-13\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-13\lt x\le5\)
Graph: Open circle at -13 and closed circle at 5. Shade between them.
\(\,\,\,\,\,\,-2\le\frac{x+1}{-3}\lt4\)
Multiply all three parts by -3 and reverse both inequality signs.
\(\,\,\,\,\,\,6\ge x+1\gt-12\)
Subtract 1 from all three parts.
\(\,\,\,\,\,\,5\ge x\gt-13\)
Rewrite in increasing order.
\(\,\,\,\,\,\,-13\lt x\le5\)
Graph: Open circle at -13 and closed circle at 5. Shade between them.
\(\textbf{11)}\) \(2x+3\lt11\) and \(x-1\ge-2\)The answer is \(-1\le x\lt4\)
Solve the first inequality.
\(\,\,\,\,\,\,2x+3\lt11\)
\(\,\,\,\,\,\,2x\lt8\)
\(\,\,\,\,\,\,x\lt4\)
Solve the second inequality.
\(\,\,\,\,\,\,x-1\ge-2\)
\(\,\,\,\,\,\,x\ge-1\)
Find the overlap because this is an “and” inequality.
\(\,\,\,\,\,\,-1\le x\lt4\)
Graph: Closed circle at -1 and open circle at 4. Shade between them.
Solve the first inequality.
\(\,\,\,\,\,\,2x+3\lt11\)
\(\,\,\,\,\,\,2x\lt8\)
\(\,\,\,\,\,\,x\lt4\)
Solve the second inequality.
\(\,\,\,\,\,\,x-1\ge-2\)
\(\,\,\,\,\,\,x\ge-1\)
Find the overlap because this is an “and” inequality.
\(\,\,\,\,\,\,-1\le x\lt4\)
Graph: Closed circle at -1 and open circle at 4. Shade between them.
\(\textbf{12)}\) \(3x-5\le4\) or \(2x+1\gt11\)The answer is \(x\le3\) or \(x\gt5\)
Solve the first inequality.
\(\,\,\,\,\,\,3x-5\le4\)
\(\,\,\,\,\,\,3x\le9\)
\(\,\,\,\,\,\,x\le3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x+1\gt11\)
\(\,\,\,\,\,\,2x\gt10\)
\(\,\,\,\,\,\,x\gt5\)
Combine using “or.”
\(\,\,\,\,\,\,x\le3\) or \(x\gt5\)
Graph: Closed circle at 3, shading left. Open circle at 5, shading right.
Solve the first inequality.
\(\,\,\,\,\,\,3x-5\le4\)
\(\,\,\,\,\,\,3x\le9\)
\(\,\,\,\,\,\,x\le3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x+1\gt11\)
\(\,\,\,\,\,\,2x\gt10\)
\(\,\,\,\,\,\,x\gt5\)
Combine using “or.”
\(\,\,\,\,\,\,x\le3\) or \(x\gt5\)
Graph: Closed circle at 3, shading left. Open circle at 5, shading right.
\(\textbf{13)}\) \(-2x+4\gt10\) or \(3x-6\ge9\)The answer is \(x\lt-3\) or \(x\ge5\)
Solve the first inequality.
\(\,\,\,\,\,\,-2x+4\gt10\)
\(\,\,\,\,\,\,-2x\gt6\)
Divide by -2 and reverse the inequality sign.
\(\,\,\,\,\,\,x\lt-3\)
Solve the second inequality.
\(\,\,\,\,\,\,3x-6\ge9\)
\(\,\,\,\,\,\,3x\ge15\)
\(\,\,\,\,\,\,x\ge5\)
Combine using “or.”
\(\,\,\,\,\,\,x\lt-3\) or \(x\ge5\)
Graph: Open circle at -3, shading left. Closed circle at 5, shading right.
Solve the first inequality.
\(\,\,\,\,\,\,-2x+4\gt10\)
\(\,\,\,\,\,\,-2x\gt6\)
Divide by -2 and reverse the inequality sign.
\(\,\,\,\,\,\,x\lt-3\)
Solve the second inequality.
\(\,\,\,\,\,\,3x-6\ge9\)
\(\,\,\,\,\,\,3x\ge15\)
\(\,\,\,\,\,\,x\ge5\)
Combine using “or.”
\(\,\,\,\,\,\,x\lt-3\) or \(x\ge5\)
Graph: Open circle at -3, shading left. Closed circle at 5, shading right.
\(\textbf{14)}\) \(4x+1\ge13\) and \(2x-3\le9\)The answer is \(3\le x\le6\)
Solve the first inequality.
\(\,\,\,\,\,\,4x+1\ge13\)
\(\,\,\,\,\,\,4x\ge12\)
\(\,\,\,\,\,\,x\ge3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x-3\le9\)
\(\,\,\,\,\,\,2x\le12\)
\(\,\,\,\,\,\,x\le6\)
Find the overlap.
\(\,\,\,\,\,\,3\le x\le6\)
Graph: Closed circles at 3 and 6. Shade between them.
Solve the first inequality.
\(\,\,\,\,\,\,4x+1\ge13\)
\(\,\,\,\,\,\,4x\ge12\)
\(\,\,\,\,\,\,x\ge3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x-3\le9\)
\(\,\,\,\,\,\,2x\le12\)
\(\,\,\,\,\,\,x\le6\)
Find the overlap.
\(\,\,\,\,\,\,3\le x\le6\)
Graph: Closed circles at 3 and 6. Shade between them.
\(\textbf{15)}\) \(x-4\lt2\) and \(2x+1\gt15\)The answer is No Solution
Solve the first inequality.
\(\,\,\,\,\,\,x-4\lt2\)
\(\,\,\,\,\,\,x\lt6\)
Solve the second inequality.
\(\,\,\,\,\,\,2x+1\gt15\)
\(\,\,\,\,\,\,2x\gt14\)
\(\,\,\,\,\,\,x\gt7\)
Combine using “and.”
\(\,\,\,\,\,\,x\lt6\) and \(x\gt7\)
These solution sets do not overlap.
The answer is No Solution.
Graph: No points are shaded in the combined solution.
Solve the first inequality.
\(\,\,\,\,\,\,x-4\lt2\)
\(\,\,\,\,\,\,x\lt6\)
Solve the second inequality.
\(\,\,\,\,\,\,2x+1\gt15\)
\(\,\,\,\,\,\,2x\gt14\)
\(\,\,\,\,\,\,x\gt7\)
Combine using “and.”
\(\,\,\,\,\,\,x\lt6\) and \(x\gt7\)
These solution sets do not overlap.
The answer is No Solution.
Graph: No points are shaded in the combined solution.
\(\textbf{16)}\) \(x+2\ge5\) or \(2x-1\lt9\)The answer is All Real Numbers
Solve the first inequality.
\(\,\,\,\,\,\,x+2\ge5\)
\(\,\,\,\,\,\,x\ge3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x-1\lt9\)
\(\,\,\,\,\,\,2x\lt10\)
\(\,\,\,\,\,\,x\lt5\)
Combine using “or.”
\(\,\,\,\,\,\,x\ge3\) or \(x\lt5\)
Every real number satisfies at least one inequality.
The answer is All Real Numbers.
Graph: Shade the entire number line.
Solve the first inequality.
\(\,\,\,\,\,\,x+2\ge5\)
\(\,\,\,\,\,\,x\ge3\)
Solve the second inequality.
\(\,\,\,\,\,\,2x-1\lt9\)
\(\,\,\,\,\,\,2x\lt10\)
\(\,\,\,\,\,\,x\lt5\)
Combine using “or.”
\(\,\,\,\,\,\,x\ge3\) or \(x\lt5\)
Every real number satisfies at least one inequality.
The answer is All Real Numbers.
Graph: Shade the entire number line.
\(\textbf{17)}\) \(-3\lt\frac{2x-1}{5}\le3\)The answer is \(-7\lt x\le8\)
\(\,\,\,\,\,\,-3\lt\frac{2x-1}{5}\le3\)
Multiply all three parts by 5.
\(\,\,\,\,\,\,-15\lt2x-1\le15\)
Add 1 to all three parts.
\(\,\,\,\,\,\,-14\lt2x\le16\)
Divide all three parts by 2.
\(\,\,\,\,\,\,-7\lt x\le8\)
Graph: Open circle at -7 and closed circle at 8. Shade between them.
\(\,\,\,\,\,\,-3\lt\frac{2x-1}{5}\le3\)
Multiply all three parts by 5.
\(\,\,\,\,\,\,-15\lt2x-1\le15\)
Add 1 to all three parts.
\(\,\,\,\,\,\,-14\lt2x\le16\)
Divide all three parts by 2.
\(\,\,\,\,\,\,-7\lt x\le8\)
Graph: Open circle at -7 and closed circle at 8. Shade between them.
\(\textbf{18)}\) \(2(x-1)\lt x+3\) and \(3(x+2)\ge2x+1\)The answer is \(-5\le x\lt5\)
Solve the first inequality.
\(\,\,\,\,\,\,2(x-1)\lt x+3\)
\(\,\,\,\,\,\,2x-2\lt x+3\)
\(\,\,\,\,\,\,x-2\lt3\)
\(\,\,\,\,\,\,x\lt5\)
Solve the second inequality.
\(\,\,\,\,\,\,3(x+2)\ge2x+1\)
\(\,\,\,\,\,\,3x+6\ge2x+1\)
\(\,\,\,\,\,\,x+6\ge1\)
\(\,\,\,\,\,\,x\ge-5\)
Find the overlap because this is an “and” inequality.
\(\,\,\,\,\,\,-5\le x\lt5\)
Graph: Closed circle at -5 and open circle at 5. Shade between them.
Solve the first inequality.
\(\,\,\,\,\,\,2(x-1)\lt x+3\)
\(\,\,\,\,\,\,2x-2\lt x+3\)
\(\,\,\,\,\,\,x-2\lt3\)
\(\,\,\,\,\,\,x\lt5\)
Solve the second inequality.
\(\,\,\,\,\,\,3(x+2)\ge2x+1\)
\(\,\,\,\,\,\,3x+6\ge2x+1\)
\(\,\,\,\,\,\,x+6\ge1\)
\(\,\,\,\,\,\,x\ge-5\)
Find the overlap because this is an “and” inequality.
\(\,\,\,\,\,\,-5\le x\lt5\)
Graph: Closed circle at -5 and open circle at 5. Shade between them.
\(\textbf{19)}\) \(0.5x+2\le5\) or \(0.25x-1\gt2\)The answer is \(x\le6\) or \(x\gt12\)
Solve the first inequality.
\(\,\,\,\,\,\,0.5x+2\le5\)
\(\,\,\,\,\,\,0.5x\le3\)
\(\,\,\,\,\,\,x\le6\)
Solve the second inequality.
\(\,\,\,\,\,\,0.25x-1\gt2\)
\(\,\,\,\,\,\,0.25x\gt3\)
\(\,\,\,\,\,\,x\gt12\)
Combine using “or.”
\(\,\,\,\,\,\,x\le6\) or \(x\gt12\)
Graph: Closed circle at 6, shading left. Open circle at 12, shading right.
Solve the first inequality.
\(\,\,\,\,\,\,0.5x+2\le5\)
\(\,\,\,\,\,\,0.5x\le3\)
\(\,\,\,\,\,\,x\le6\)
Solve the second inequality.
\(\,\,\,\,\,\,0.25x-1\gt2\)
\(\,\,\,\,\,\,0.25x\gt3\)
\(\,\,\,\,\,\,x\gt12\)
Combine using “or.”
\(\,\,\,\,\,\,x\le6\) or \(x\gt12\)
Graph: Closed circle at 6, shading left. Open circle at 12, shading right.
\(\textbf{20)}\) \(2\le3x-1\lt2x+7\)The answer is \(1\le x\lt8\)
\(\,\,\,\,\,\,2\le3x-1\lt2x+7\)
Separate into two inequalities.
\(\,\,\,\,\,\,2\le3x-1\)
\(\,\,\,\,\,\,3\le3x\)
\(\,\,\,\,\,\,1\le x\)
Now solve the second inequality.
\(\,\,\,\,\,\,3x-1\lt2x+7\)
\(\,\,\,\,\,\,x-1\lt7\)
\(\,\,\,\,\,\,x\lt8\)
Both inequalities must be true.
\(\,\,\,\,\,\,1\le x\lt8\)
Graph: Closed circle at 1 and open circle at 8. Shade between them.
\(\,\,\,\,\,\,2\le3x-1\lt2x+7\)
Separate into two inequalities.
\(\,\,\,\,\,\,2\le3x-1\)
\(\,\,\,\,\,\,3\le3x\)
\(\,\,\,\,\,\,1\le x\)
Now solve the second inequality.
\(\,\,\,\,\,\,3x-1\lt2x+7\)
\(\,\,\,\,\,\,x-1\lt7\)
\(\,\,\,\,\,\,x\lt8\)
Both inequalities must be true.
\(\,\,\,\,\,\,1\le x\lt8\)
Graph: Closed circle at 1 and open circle at 8. Shade between them.
See Related Pages\(\)
\(\bullet\text{ Compound Inequality Calculator}\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Solving Inequalities Addition and Subtraction}\)
\(\,\,\,\,\,\,\,\,4+x\ge12…\)
\(\bullet\text{ Solving Inequalities Multiplication and Division}\)
\(\,\,\,\,\,\,\,\,4x\lt16…\)
\(\bullet\text{ Solving Multi-step Inequalities}\)
\(\,\,\,\,\,\,\,\,18-4x\lt2…\)
\(\bullet\text{ Andymath Homepage}\)
