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Exercises 1 To solve for z we need to isolate it by subtracting 3 from both sides of the equation. We want to subtract 3 because addition and subtraction are inverse operations - they cancel each other out. This is allowed by the Subtraction Property of Equality. z+3=-6LHS−3=RHS−3z+3−3=-6−3Subtract termsz=-9 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. z+3=-6z=-9-9+3=?-6Add terms-6=-6 An identity occurs when the left-hand side and the right-hand side of an equation are equal to one another. Since substituting z=-9 resulted in an identity, our solution is correct.
Exercises 2 To solve the equation, we have to isolate the variable. Notice that the coefficient of t is -0.2. To undo this multiplication we will use the Division Property of Equality to divide both sides of the equation by -0.2. 2.6=-0.2tLHS/(-0.2)=RHS/(-0.2)-0.22.6​=-0.2-0.2t​Put minus sign in front of fraction-0.22.6​=-0.2-0.2t​Calculate quotient-13=tRearrange equationt=-13 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. 2.6=-0.2tt=-132.6=?-0.2⋅(-13)-a(-b)=a⋅b2.6=?0.2⋅13Multiply2.6=2.6 An identity occurs when the left-hand side and the right-hand side of an equation are equal to one another. Since substituting t=-13 resulted in an identity, our solution is correct.
Exercises 3 To solve for n, we need to isolate it by multiplying both sides of the equation by -5. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. -5n​=-2Put minus sign in denominator-5n​=-2LHS⋅-5=RHS⋅-5-5n​⋅-5=-2⋅-5-5a​⋅-5=an=-2⋅-5Multiplyn=10 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. -5n​=-2n=10-510​=?-2Calculate quotient-2=-2 An identity occurs when the left-hand side and the right-hand side of an equation are equal to one another. Since substituting n=10 resulted in an identity, our solution is correct.
Exercises 4 To solve the equation, we first have to isolate 3y by subtracting 11 from both sides. 3y+11=-16LHS−11=RHS−113y+11−11=-16−11Subtract term3y=-27 The next step is to isolate y on the left-hand side by dividing both sides of the equation by 3. 3y=-27LHS/3=RHS/333y​=3-27​Put minus sign in front of fraction33y​=-327​Calculate quotienty=-9 Therefore, y=-9 is the solution to the equation. We can check our solution by substituting it into the original equation. 3y+11=-16y=-93(-9)+11=?-16a(-b)=-a⋅b-3⋅9+11=?-16Multiply-27+11=?-16Add terms-16=-16 Since the left-hand side and right-hand side are equal, we have the correct solution.
Exercises 5 To solve the equation, we first have to isolate -b. We can do this by subtracting 1 from both sides. 6=1−b LHS−1=RHS−1 Rewrite 1−b as -b+16=-b+1LHS−1=RHS−16−1=-b+1−1Subtract terms 5=-bLHS⋅-1=RHS⋅-1-5=bRearrange equationb=-5 Therefore, b=-5 is the solution to the equation. We can check our solution by substituting it into the original equation. 6=1−bb=-56=?1−(-5)a−(-b)=a+b6=?1+5Add terms6=6 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 6 We can solve the equation by combining like terms and then isolating n. n+5n+7=43Add terms6n+7=43LHS−7=RHS−76n+7−7=43−7Subtract terms6n=36LHS/6=RHS/666n​=636​Calculate quotientn=6 Let's check our answer by substituting n=6 into the original equation. n+5n+7=43n=66+5(6)+7=?43 Simplify LHS Multiply6+30+7=?43Add terms 43=43 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 7 We have been asked to solve the equation -4(2z+6)−12=4. We can begin by distributing the -4 to the terms in the parentheses. Then we will continue isolating the variable z. -4(2z+6)−12=4Distribute -4(-4)2z+(-4)6−12=4(-a)b=-ab-8z+(-24)−12=4a+(-b)=a−b-8z−24−12=4Subtract terms-8z−36=4LHS+36=RHS+36-8z−36+36=4+36Add terms-8z=40LHS/(-8)=RHS/(-8)-8-8z​=-840​Calculate quotientz=-5 Now we have that z=-5. Let's check our answer by substituting z=-5 into the original equation and following the order of operations. -4(2z+6)−12=4z=-5-4(2(-5)+6)−12=?4 Simplify LHS a(-b)=-a⋅b-4(-10+6)−12=?4Add terms-4(-4)−12=?4-a(-b)=a⋅b16−12=?4Subtract terms 4=4 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 8 To solve an equation, we need to isolate the variable. Before we can do that, we must distribute on the left-hand side. 23​(x−2)−5=19Distribute 23​23​⋅x−23​⋅2−5=192a​⋅2=a23​⋅x−3−5=19ca​⋅b=ca⋅b​23x​−3−5=19Subtract term23x​−8=19LHS+8=RHS+823x​=27LHS⋅32​=RHS⋅32​32​⋅23x​=32​⋅27ba​⋅ab​=1x=32​⋅27ca​⋅b=ca⋅b​x=354​Calculate quotientx=18 The solution to the equation is x=18. We can check our solution by substituting it into the original equation and simplifying. 23​(x−2)−5=19x=1823​(18−2)−5=?19 Simplify LHS & RHS Subtract term23​⋅16−5=?19ca​⋅b=ca⋅b​248​−5=?19Calculate quotient24−5=?19Subtract term 19=19 Because the left-hand side and right-hand sides are equal, we know that our solution is correct.
Exercises 9 We need to isolate the variable w on one side to solve the equation. Notice that 5⋅7=35 so when it comes time to eliminate the fraction we will multiply both sides by 35. 6=51​w+71​w−4LHS+4=RHS+410=51​w+71​wLHS⋅35=RHS⋅35350=7w+5wAdd terms350=12wLHS/12=RHS/1212350​=wRearrange equationw=12350​ba​=b/2a/2​w=6175​
Exercises 10 It is given in the exercise that the sum of all interior angles in a triangle equals 180∘. We can thus write an equation that adds all of the angles together and is equal to 180∘. 110∘+5x∘+2x∘=180∘ By solving this equation for x we will be able to find the measure of the two missing angles. 110+5x+2x=180Add terms110+7x=180LHS−110=RHS−110110+7x−110=180−110Subtract terms7x=70LHS/7=RHS/777x​=770​ca⋅b​=ca​⋅b77​x=770​Calculate quotientx=10 We know now that the value of x is 10 and we also know the value of one of the angles, 110∘. To find the value of the two missing angles we substitute 10 for x in 5x and in 2x. 5x∘⇒5⋅10∘=50∘2x∘⇒2⋅10∘=20∘​ Therefore, the two missing angles are 50∘ and 20∘.
Exercises 11 It is given that the sum of all interior angles in the given pentagon is 540∘. We can thus write an equation that adds all of the angles together and is equal to 540∘. (x−30)+(x−30)+(x−30)+x+x=540 Solving this equation for x will give us the measure of the unknown angles. (x−30)+(x−30)+(x−30)+x+x=540Associative Property of Additionx−30+x−30+x−30+x+x=540Add and subtract terms5x−90=540LHS+90=RHS+905x=630LHS/5=RHS/5x=126 The the unknown angles x measure 126∘ and the unknown angles x−30 measure 96∘.
Exercises 12 To solve the equation, we have to isolate n on one side. To begin, we will move the variable terms to one side. 3n−3=4n+1LHS−3n=RHS−3n3n−3−3n=4n+1−3nSubtract terms-3=n+1LHS−1=RHS−1-3−1=n+1−1Subtract terms-4=nRearrange equationn=-4 The solution to the equation is n=-4.
Exercises 13 To solve the equation 5(1+x)=5x+5, we need to isolate the variable. In the given equation, notice that there are two variable terms. We will need to move them to one side to solve. Before we can do that, we must distribute on the left-hand side. 5(1+x)=5x+5Distribute 55⋅1+5⋅x=5x+5Multiply5+5x=5x+5LHS−5x=RHS−5x5+5x−5x=5x+5−5xSubtract terms5=5 We have arrived to 5=5, which is an identity. Whenever we arrive at an identity, it means the equation is true for all values of x. Therefore, the solution set for this equation is all real numbers.
Exercises 14 We will solve this exercise by using the Distributive Property twice. Let's start by distributing the 3 to the terms inside the parentheses on the left-hand side. 3(n+4)=21​(6n+4)Distribute 33⋅n+3⋅4=21​(6n+4)Distribute 21​3⋅n+3⋅4=21​⋅6n+21​⋅4ca​⋅b=ca⋅b​3⋅n+3⋅4=26n​+24​Multiply3n+12=26n​+24​Calculate quotient3n+12=3n+2LHS−3n=RHS−3n12​=2 Since 12 cannot be equal to 2, we have reached a contradiction. This means that there is no solution to this equation.
Exercises 15 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint on a number line. In this exercise, the equation ∣y+3∣=17 means that the value of the expression y+3 is 17 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣y+3∣=17⇒y+3=-17y+3=-17​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣y+3∣=17y+3≥0: y+3=17y+3<0: y+3=-17​(I)(II)​y+3=17y+3=-17​(I)(II)​(I), (II):  LHS−3=RHS−3y1​=14y2​=-20​ Both y=14 and y=-20 are solutions to the given equation. By substituting y=14 and y=-20 into the equation and evaluating, we can check if the solutions are correct. First, let's check y=14. ∣y+3∣=17y=14∣14+3∣=?17Add terms∣17∣=?17∣17∣=1717=17 Because the left-hand and right-hand sides are equal, we know that the solution is correct. Now we will check the second solution by substituting -20 for y. ∣y+3∣=17y=-20∣-20+3∣=?17Add terms∣-17∣=?17∣-17∣=1717=17 Therefore, both solutions are correct.
Exercises 16 Before we begin, let's isolate the absolute value on the left-hand side. -2∣5w−7∣+9=-7LHS−9=RHS−9-2∣5w−7∣+9−9=-7−9Subtract terms-2∣5w−7∣=-16LHS/-2=RHS/-2∣5w−7∣=8 An absolute value is always a non-negative number because it measures the expression's distance from zero on a number line. In this exercise, the equation ∣5w−7∣=8 means that the value of the expression 5w−7 is 8 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣5w−7∣=8⇒5w−7=-85w−7=-8​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣5w−7∣=85w−7≥0: 5w−7=85w−7<0: 5w−7=-8​(I)(II)​5w−7=85w−7=-8​(I)(II)​(I), (II):  LHS+7=RHS+75w=155w=-1​(I), (II):  LHS/5=RHS/5w=3w=-51​​ Both w=3 and w=-51​ are solutions to the given equation. By substituting w=3 and w=-51​ into the equation and evaluating, we can check if the solutions are correct. First, let's check w=3. ∣5w−7∣=8w=3∣5(3)−7∣=?8Multiply∣15−7∣=?8Subtract terms∣8∣=?8∣8∣=88=8 Because the left-hand and right-hand sides are equal, we know that the solution is correct. Now we will check the second solution by substituting -51​ for w. ∣5w−7∣=8w=-51​∣5(-51​)−7∣=?8Put minus sign in numerator∣5(5-1​)−7∣=?85a​⋅5=a∣-1−7∣=?8Subtract terms∣-8∣=?8∣-8∣=88=8 Both sides of the equation are equal. Therefore, both solutions are correct.
Exercises 17 When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=∣cx+d∣​ then we can have either ax+b=cx+dorax+b=-(cx+d)​ To solve the given absolute value equation, we will write two equations like above when we remove the absolute value. ∣x−2∣=∣4+x∣x−2≥0: x−2=(4+x)x−2<0: x−2=-(4+x)​(I)(II)​x−2=4+xx−2=-(4+x)​(I)(II)​ Solve for x (II):  Remove parentheses and change signsx−2=4+xx−2=-4−x​(I): LHS−x=RHS−x-2​=4x−2=-4−x​(II): LHS+x=RHS+x-2​=42x−2=-4​(II): LHS+2=RHS+2-2​=42x=-2​(II): LHS/2=RHS/2 -2​=4x=-1​ Notice that when solving the first equation we encountered a contradiction. Thus, the first equation has no solution. The solution to Equation (II) is x=-1. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. ∣x−2∣=∣4+x∣x=-1∣-1−2∣=?∣4+-1∣ Simplify equation Add and subtract terms∣-3∣=?∣3∣∣-3∣=3 3=3 x=-1 is not extraneous.
Exercises 18 An absolute value equation takes on the following form: ∣v−midpoint∣=distance to midpoint​ To write an absolute value equation that models the situation, we can begin by thinking about the given minimum and maximum lengths as solutions to the equation. By plotting these points on a number line we can determine the midpoint and the distance from each point to the midpoint.From the number line, we can see that the midpoint between 74 and 95 is 84.5, and that the distance from both values to the midpoint is 10.5. We can then write the following equation. ∣v−84.5∣=10.5​ We can solve the equation we've created to make sure it has the desired solutions. ∣v−84.5∣=10.5v−84.5≥0: v−84.5=10.5v−84.5<0: v−84.5=-10.5​(I)(II)​v−84.5=10.5v−84.5=-10.5​(I)(II)​(I), (II):  LHS+84.5=RHS+84.5v=95v=74​ Since our created equation yielded the values 95 and 74, we can conclude that it is correct.
Exercises 19 An equation that has two or more variables is called a literal equation. To rewrite a literal equation, we need to solve for one variable in terms of the other variable(s). In this exercise we are asked to solve for y. 2x−4y=20LHS−2x=RHS−2x2x−4y−2x=20−2xSubtract terms-4y=20−2xLHS/(-4)=RHS/(-4)-4-4y​=-420−2x​ca⋅b​=ca​⋅b-4-4​y=-420−2x​-b-a​=ba​y=-420−2x​ Simplify RHS Write as a difference of fractionsy=-420​−-42x​ca⋅b​=ca​⋅by=-420​−-42​xCalculate quotienty=-5−-42​xPut minus sign in front of fractiony=-5−(-42​)x-(-a)=ay=-5+42​xba​=b/2a/2​ y=-5+21​x We have isolated y on the left-hand side of the equation and simplified the expression on the right-hand side.
Exercises 20 To solve for y, we have to isolate y on one side of the equation using inverse operations. 8x−3=5+4yLHS−5=RHS−58x−8=4yLHS/4=RHS/448x−8​=44y​Write as a difference of fractions48x​−48​=44y​Calculate quotient2x−2=yRearrange equationy=2x−2
Exercises 21 We solve the equation for y, by isolating y on one side. We can use the Properties of Equality just like we would solve any equation. To begin, we will factor y out of the expression on the right-hand side of the equation. a=9y+3yxFactor out ya=y(9+3x)LHS/9+3x=RHS/9+3x9+3xa​=yRearrange equationy=9+3xa​
Exercises 22
Exercises 23