#### Quiz

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##### Sections
###### Exercises
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Exercises 1 To solve for x, we need to isolate it by subtracting 9 from both sides of the equation. We want to subtract 9 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Subtraction Property of Equality. x+9=7LHS−9=RHS−9x+9−9=7−9Subtract termsx=-2 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. x+9=7x=-2-2+9=?7Add terms7=7 Since substituting x=-2 resulted in an identity, our solution is correct.
Exercises 2 To solve for z, we need to isolate it by adding 3.8 to both sides of the equation. We want to add 3.8 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. 8.6=z−3.8LHS+3.8=RHS+3.88.6+3.8=z−3.8+3.8Add terms12.4=zRearrange equationz=12.4 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 8.6=z−3.8z=12.48.6=?12.4−3.8Subtract term8.6=8.6 Since substituting z=12.4 resulted in an identity, our solution is correct.
Exercises 3 To solve for r, we need to isolate it by dividing both sides of the equation by -12. We want to divide because multiplication and division are inverse operations. This is allowed by the Division Property of Equality. 60=-12rLHS/(-12)=RHS/(-12)-1260​=-12-12r​Calculate quotient-5=rRearrange equationr=-5 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 60=-12rr=-560=?-12(-5)Multiply60=60 Since substituting r=-5 resulted in an identity, our solution is correct.
Exercises 4 To solve for p, we first need to isolate 3p by multiplying both sides of the equation by 4. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. 43​p=18ca​⋅b=ca⋅b​43p​=18LHS⋅4=RHS⋅443p​⋅4=18⋅44a​⋅4=a3p=18⋅4Multiply3p=72 To continue isolating p, we will divide both sides of the equation by 3. We are allowed to do this by the Division Property of Equality. 3p=72LHS/3=RHS/333p​=372​Calculate quotientp=24 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 43​p=18p=2443​(24)=?18ca​⋅b=ca⋅b​43⋅24​=?18Multiply472​=?18Calculate quotient18=18 Since substituting p=24 resulted in an identity, our solution is correct.
Exercises 5 To solve the equation, we first isolate 2m by adding 3 to both sides. 2m−3=13LHS+3=RHS+32m−3+3=13+3Add terms2m=16 The next step is to isolate m on the left-hand side. Using the Division Property of Equality to divide both sides of the equation by 2, we can isolate m. 2m=16LHS/2=RHS/222m​=216​Calculate quotientm=8 We can check our solution by substituting it into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 2m−3=13m=82(8)−3=?13Multiply16−3=?13Subtract term13=13 Because the left and right-hand sides of the equation are the same, we know that m=8 is the correct solution.
Exercises 6 To solve the equation, we first have to isolate -v. According to the Subtraction Property of Equality, we can do this by subtracting 10 from both sides. 5=10−vLHS−10=RHS−105−10=10−10−vSubtract terms-5=-v We continue isolating v by dividing both sides by -1. -5=-vLHS/(-1)=RHS/(-1)-1-5​=-1-v​-b-a​=ba​5=vRearrange equationv=5 We can check our solution by substituting it into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 5=10−vv=55=?10−5Subtract term5=5 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 7 We can solve the equation by combining like terms and then isolating w. 5=7w+8w+2Add terms5=15w+2LHS−2=RHS−25−2=15w+2−2Subtract terms3=15wLHS/15=RHS/15153​=wba​=b/3a/3​51​=wRearrange equationw=51​ Let's check our solution by substituting w=51​ into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 5=7w+8w+2w=51​5=?7(51​)+8(51​)+2a⋅b1​=ba​5=?57​+58​+2 Simplify RHS Add fractions5=?515​+2Calculate quotient5=?3+2Add terms 5=5 Because both sides are equal, we know that our answer is correct.
Exercises 8 We can solve the equation by combining like terms and then isolating a. -21a+28a−6=-10.2Add terms7a−6=-10.2LHS+6=RHS+67a−6+6=-10.2+6Add terms7a=-4.2LHS/7=RHS/777a​=7-4.2​Use a calculatora=-0.6 Let's check our answer by substituting p=-0.6 into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. -21a+28a−6=-10.2a=-0.6-21(-0.6)+28(-0.6)−6=?-10.2-a(-b)=a⋅b12.6+28(-0.6)−6=?-10.2a(-b)=-a⋅b12.6−16.8−6=?-10.2Subtract terms-10.2=-10.2 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 9 To begin solving this equation, we distribute the -3 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 2k−3(2k−3)=45Distribute -32k−6k+9=45Subtract term-4k+9=45LHS−9=RHS−9-4k+9−9=45−9Subtract terms-4k=36LHS/(-4)=RHS/(-4)-4-4k​=-436​Calculate quotientk=-9 Therefore, k=-9 is the solution to the equation. We can check our answer by substituting it back into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 2k−3(2k−3)=45k=-92(-9)−3(2(-9)−3)=?45 Simplify LHS a(-b)=-a⋅b-18−3(-18−3)=?45Subtract term-18−3(-21)=?45-a(-b)=a⋅b -18+63=?45Subtract term45=45 Because both sides are equal, we know that our answer is correct.
Exercises 10 To begin solving this equation, we distribute the 51​ to each term in the parentheses on the right-hand side. Then we can continue solving by using the Properties of Equality. 68=51​(20x+50)+2Distribute 51​68=4x+10+2Add terms68=4x+12LHS−12=RHS−1268−12=4x+12−12Subtract term56=4xLHS/4=RHS/4456​=44x​Calculate quotient14=xRearrange equationx=14 Therefore, x=14 is the solution to the equation. We can check our answer by substituting it back into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same. 68=51​(20x+50)+2x=1468=?51​(20(14)+50)+2Multiply68=?51​(280+50)+2Add terms68=?51​⋅330+2b1​⋅a=ba​68=?5330​+2Calculate quotient68=?66+2Add terms68=68 Because both sides are equal, we know that our answer is correct.
Exercises 11 To solve the equation, we have to isolate c on one side. To begin, we will move the variable terms to one side. 3c+1=c+1LHS−c=RHS−c3c+1−c=c+1−cSubtract terms2c+1=1LHS−1=RHS−12c+1−1=1−1Subtract terms2c=0LHS/2=RHS/222c​=20​Calculate quotientc=0 The solution to the equation is c=0.
Exercises 12 To solve the equation, we have to isolate n on one side. We accomplish this by moving the variable terms to the left-hand side and the constant terms to the right-hand side. -8−5n=64+3nLHS−3n=RHS−3n-8−5n−3n=64+3n−3nSubtract terms-8−8n=64LHS+8=RHS+8-8−8n+8=64+8Add terms-8n=72 Finally, we need to divide both sides by -8 to fully isolate n. -8n=72LHS/(-8)=RHS/(-8)-8-8n​=-872​Calculate quotientn=-9 Therefore, n=-9 is the solution to the equation.
Exercises 13 First, using the Distributive Property, we can simplify the left-hand side. 2(8q−5)=4qDistribute 216q−10=4q Now, we can continue to solve using the Properties of Equality. 16q−10=4qLHS+10=RHS+1016q−10+10=4q+10Add terms16q=4q+10LHS−4q=RHS−4q16q−4q=4q+10−4qSubtract terms12q=10LHS/12=RHS/121212q​=1210​Calculate quotientq=1210​ba​=b/2a/2​q=65​ The solution to the equation is q=65​.
Exercises 14 On both sides of the equation, we have products that can be simplified using the Distributive Property. Let's do that first. 9(y−4)−7y=5(3y−2)Distribute 9 & 59y−36−7y=15y−10Subtract term2y−36=15y−10 Now, we can continue solving for y using inverse operations. We need to move all the constants to the right and the variable terms to the left. 2y−36=15y−10LHS+36=RHS+362y−36+36=15y−10+36Add terms2y=15y+26LHS−15y=RHS−15y2y−15y=15y+26−15ySubtract term-13y=26LHS/(-13)=RHS/(-13)-13-13y​=-1326​Calculate quotienty=-2 The solution to the equation is y=-2.
Exercises 15 To begin solving this equation, we can distribute the 4 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 4(g+8)=7+4gDistribute 44g+32=7+4gLHS−4g=RHS−4g4g+16−4g=7+4g−4gSubtract terms32≠7 Simplifying the equation resulted in a contradiction. Therefore, the equation has no solution.
Exercises 16 On both sides of the equation, we have products that can be simplified using the Distributive Property. -4(-5h−4)=2(10h+8)Distribute -420h+16=2(10h+8)Distribute 220h+16=20h+16 Now, we can continue to solve using the Properties of Equality. 20h+16=20h+16LHS−20h=RHS−20h20h+16−20h=20h+16−20hSubtract terms16=16 Simplifying the given equation resulted in an identity, both sides of the equation are exactly the same. Therefore, the equation has infinitely many solutions.
Exercises 17 The miles you are from a thunderstorm is calculated by the seconds s that pass between seeing lightning and hearing the clap, divided by 5. miles away =5s​ If a storm is 2 miles away, we can substitute that into the equation above and solve for the number of seconds we would need to count. miles away =5s​Substitute values2=5s​LHS⋅5=RHS⋅510=sRearrange equations=10 When we can count to 10 between seeing the lightning and hearing the thunder, a storm is approximately 2 miles away.
Exercises 18 We need the end spaces, the posters, and the spaces between the posters to add to 15 feet, the full length of the wall. We can add the given lengths, the lengths of the end spaces and the posters, and consider them to be a singular constant when writing an equation. 3+3+2+2+2=12 feet Let the spaces between the posters be s. We know that our 15 ft wall needs to equal the length of 12 posters plus 2 times the unknown spaces between the posters. 15=12+2s Let's solve for s now! 15=12+2sLHS−12=RHS−123=2sLHS/2=RHS/223​=sWrite as a decimal1.5=sRearrange termss=1.5 The spaces between the posters should be 1.5 feet each in order to have everything evenly spaced along the wall.
Exercises 19 aTo find after how many hours the cost of the studios will be the same, we need to write an equation for the cost of each studio and then set them equal to one another. Studio A has a flat fee of \$10 for the vase and then costs \$8 per hour h. A=10+8h Studio B has a flat fee of \$16 for the vase and then costs \$6 per hour h. B=16+6h Now, we can find when Studio A= Studio B. A=BA=10+8h, B=16+6h10+8h=16+6hLHS−10=RHS−108h=6+6hLHS−6h=RHS−6h2h=6LHS/2=RHS/2h=3 If you use the Studio for 3 hours, both will cost the same.bLet's suppose that Studio B increases their price by \$2 per hour. This would change the equation for the total cost. B=16+8h Let's set the equations equal to one another again and solve to see what happens. A=BA=10+8h, B=16+8h10+8h=16+8hLHS−10=RHS−108h=6+8hLHS−8h=RHS−8h0≠6 There is no solution. This makes sense if we think about it, even without using equations. The vase at Studio B costs \$6 more than at Studio A. If the hourly rates are the exact same, Studio B will always be more expensive.