| Exercises 1 A literal equation is an equation that has two or more variables. The equation 9r+16=5π only has one variable, r. Notice that π is not a variable because its value is constant:
| Exercises 2 To begin, we will label each prompt to make discussing them easier.NameEquationPrompt
A3x+6y=24Solve for x.
B24−3x=6ySolve for x.
C6y=24−3xSolve for y in terms of x.
D24−6y=3xSolve for x in terms of y.
One possible answer
Notice that in the table above, C is the only prompt that asks to solve for y. All others ask that we solve for x. Thus, C does not belong.Another possible answer
Also, from the table, we can notice that D is the only equation that shows the x term on one side. All other equations show the x term with another term. Thus, D does not belong.|| |
| Exercises 3 To solve for y we have to isolate the variable on one side of the equation.
y−3x=13LHS+3x=RHS+3xy−3x+3x=13+3xAdd termsy=13+3x|| |
| Exercises 4 To solve for y, we have to isolate the variable on one side of the equation.
2x+y=7LHS−2x=RHS−2x2x+y−2x=7−2xSubtract termsy=7−2x|| |
| Exercises 5 To solve for y, we have to isolate the variable on one side of the equation.
2y−18x=-26LHS+18x=RHS+18x2y−18x+18x=-26+18xAdd terms2y=-26+18xLHS/2=RHS/2y=2-26+18xWrite as a sum of fractionsy=2-26+218xCalculate quotienty=-13+9x|| |
| Exercises 6 To solve for y we have to isolate the variable on one side of the equation.
20x+5y=15LHS−20x=RHS−20x20x+5y−20x=15−20xSubtract terms5y=15−20xLHS/5=RHS/5y=515−20xWrite as a difference of fractionsy=515−520xCalculate quotienty=3−4x|| |
| Exercises 7 To solve for y we have to isolate the variable on one side of the equation.
9x−y=45LHS+y=RHS+y9x−y+y=45+yAdd terms9x=45+yLHS−45=RHS−459x−45=45+y−45Subtract terms9x−45=yRearrange equationy=9x−45|| |
| Exercises 8 To solve for y we have to isolate the variable on one side of the equation.
6−3y=-6LHS+3y=RHS+3y6−3y+3y=-6+3yAdd terms6=-6+3yLHS+6=RHS+66+6=-6+3y+6Add terms12=3yLHS/3=RHS/3312=33yCalculate quotient4=yRearrange equationy=4|| |
| Exercises 9 To solve for y we have to isolate the variable on one side of the equation.
4x−5=7+4yLHS−7=RHS−74x−5−7=7+4y−7Subtract terms4x−12=4yLHS/4=RHS/444x−12=44yWrite as a difference of fractions44x−412=44yCalculate quotientx−3=yRearrange equationy=x−3|| |
| Exercises 10 To solve for y we have to isolate the variable on one side of the equation.
16x+9=9y−2xLHS+2x=RHS+2x16x+9+2x=9y−2x+2xAdd terms18x+9=9yLHS/9=RHS/9918x+9=99yWrite as a sum of fractions918x+99=99yCalculate quotient2x+1=yRearrange equationy=2x+1|| |
| Exercises 11 To solve the literal equation for y, we will isolate y.
2+61y=3x+4LHS−2=RHS−261y=3x+2Multiply by 6y=6(3x+2)Distribute 6y=6⋅3x+6⋅2Multiplyy=18x+12|| |
| Exercises 12 We need to solve the given literal equation for y.
11−21y=3+6xLHS−11=RHS−1111−21y−11=3+6x−11Subtract term-21y=-8+6xca⋅b=ca⋅b-2y=-8+6xPut minus sign in denominator-2y=-8+6x
Now we can eliminate the fraction by multiplying both sides of the equation by -2.
-2y=-8+6xLHS⋅(-2)=RHS⋅(-2)-2y⋅(-2)=(-8+6x)(-2)(-2)a⋅(-2)=ay=(-8+6x)(-2)Distribute -2y=16−12x|| |
| Exercises 13 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
y=4x+8xFactor out xy=x(4+8)Add termsy=x(12)Multiplyy=12xLHS/12=RHS/1212y=1212xCalculate quotient12y=x1⋅a=a121⋅y=xca⋅b=ca⋅b121y=xRearrange equationx=121y|| |
| Exercises 14 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
m=10x−xFactor out m=x(10−1)Subtract termm=x(9)Multiplym=9xLHS/9=RHS/99m=99xCalculate quotient9m=x1⋅a=a91⋅m=xca⋅b=ca⋅b91m=xRearrange equationx=91m|| |
| Exercises 15 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
a=2x+6xzFactor out xa=x(2+6z)LHS/2+6z=RHS/2+6z2+6za=2+6zx(2+6z)ba=b/2+6za/2+6z2+6za=1x1a=a2+6za=xRearrange equationx=2+6za|| |
| Exercises 16 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
y=3bx−7xFactor out xy=x(3b−7)LHS/3b−7=RHS/3b−73b−7y=3b−7x(3b−7)ba=b/3b−7a/3b−73b−7y=1x1a=a3b−7y=xRearrange equationx=3b−7y|| |
| Exercises 17 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
y=4x+rx+6LHS−6=RHS−6y−6=4x+rx+6−6Subtract termsy−6=4x+rxFactor out xy−6=x(4+r)LHS/4+r=RHS/4+r4+ry−6=xRearrange equationx=4+ry−6|| |
| Exercises 18 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
z=8+6x−pxLHS−8=RHS−8z−8=8+6x−px−8Subtract termz−8=6x−pxFactor out xz−8=x(6−p)LHS/6−p=RHS/6−p6−pz−8=xRearrange equationx=6−pz−8|| |
| Exercises 19 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
sx+tx=rFactor out xx(s+t)=rLHS/s+t=RHS/s+tx=s+tr|| |
| Exercises 20 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
a=bx+cx+dLHS−d=RHS−da−d=bx+cxFactor out xa−d=x(b+c)LHS/b+c=RHS/b+cb+ca−d=xRearrange equationx=b+ca−d|| |
| Exercises 21 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
12−5x−4kx=yLHS−12=RHS−1212−5x−4kx−12=y−12Subtract term-5x−4kx=y−12Factor out xx(-5−4k)=y−12LHS/-5−4k=RHS/-5−4kx=-5−4ky−12|| |
| Exercises 22 We solve the equation for x, by isolating x on one side. We can use the Properties of Equality just like we would solve any equation.
x−9+2wx=yLHS+9=RHS+9x−9+2wx+9=y+9Add termsx+2wx=y+9Factor out xx(1+2w)=y+9LHS/1+2w=RHS/1+2wx=1+2wy+9|| |
| Exercises 23 aTo solve the equation for x, means we should isolate x.
C=85x+60LHS−60=RHS−60C−60=85xLHS/85=RHS/8585C−60=xRearrange equationx=85C−60bTo determine the number of ski trips we can take for 315 dollars and 485 dollars, we can substitute those values for C into the equation from Part A and solve for x.C85C−60x
If we spend 315 dollars, we can have 3 ski trips. If we spend 485 dollars, we can have 5 ski trips.|| |
| Exercises 24 aTo solve the equation for n, means we should isolate n.
d=4n−2LHS+2=RHS+2d+2=4n−2+2Add termsd+2=4nLHS/4=RHS/44d+2=xRearrange equationn=4d+2bTo determine the lengths of nails we can take for 3, 6 and 10, we can substitute those values for d into the equation from Part A and solve for n.d4d+2n
If the penny size is 3, the length of a nail is 45 in. If the penny size is 6, the length of a nail is 2 in. If the penny size is 10, the length of a nail is 3 in.|| |
| Exercises 25 The original solver of this equation decided to treat (y−x) as though it was a single unit rather than the variable y minus the variable x. To separate the variables, we need to remove the parentheses using the Distributive Property. After we have distributed, we can use the Properties of Equality to fully isolate x.
12−2x=-2(y−x)Distribute -212−2x=-2y+2xLHS+2x=RHS+2x12=-2y+4xLHS+2y=RHS+2y2y+12=4xRearrange equation4x=2y+12LHS/4=RHS/4x=21y+3|| |
| Exercises 26 In the attempted solution, an x was factored out on the right-hand side but this was done in error. When something is "factored out," it must be as though you did the Distributive Property in reverse. The second line shows x(a−3b) on the right-hand side, however,
This is not the same as ax−3b, which is what we started with in the first line. Instead, we should solve for x by first adding 3b to both sides of the equation,
10=ax−3bLHS+3b=RHS+3b10+3b=axRearrange equationax=10+3bLHS/a=RHS/ax=a10+3b|| |
| Exercises 27 To solve the given equation for C, we will isolate C.
P=R−CLHS−R=RHS−RP−R=-CChange signs-P+R=CRearrange equationC=-P+RRearrange termsC=R−P
Solving the given equation for C yields C=R−P.|| |
| Exercises 28 To solve the given equation for h, we will isolate h.
S=2πr2+2πrhLHS−2πr2=RHS−2πr2S−2πr2=2πr2+2πrh−2πr2Subtract termS−2πr2=2πrhLHS/2πr=RHS/2πr2πrS−2πr2=hRearrange equationh=2πrS−2πr2|| |
| Exercises 29 To solve the given equation for b2, we will isolate b2.
A=21h(b1+b2)LHS⋅2=RHS⋅22A=2⋅21h(b1+b2)2⋅2a=a2A=h(b1+b2)LHS/h=RHS/hh2A=hh(b1+b2)ba=b/ha/hh2A=1(b1+b2)1a=ah2A=(b1+b2)Remove parenthesesh2A=b1+b2LHS−b1=RHS−b1h2A−b1=b2Rearrange equationb2=h2A−b1|| |
| Exercises 30 To solve for v1, we have to isolate v1 on one side. We can use the Properties of Equality just like we would solve any equation.
a=tv1−v0LHS⋅t=RHS⋅tat=tv1−v0⋅tta⋅t=aat=v1−v0LHS+v0=RHS+v0at+v0=v1−v0+v0Add termsat+v0=v1Rearrange equationv1=at+v0|| |
| Exercises 31 Let's solve for C by applying inverse operations until we have isolated C
Since C is isolated on one side, we have solved the given equation for C.|| |
| Exercises 32 Let's solve for m1 by applying inverse operations.
F=G(d2m1m2)LHS/G=RHS/GGF=d2m1m2LHS⋅d2=RHS⋅d2GF⋅d2=m1m2LHS/m2=RHS/m2GF⋅d2/m2=m1ca⋅b=ca⋅bGFd2/m2=m1ba/c=b⋅caGm2Fd2=m1Rearrange equationm1=Gm2Fd2|| |
| Exercises 33 aSolving the formula for r requires the isolation of r on one side.
S=L−rLLHS−L=RHS−LS−L=-rLChange signs-S+L=rLRearrange termsL−S=rLLHS/L=RHS/LLL−S=rRearrange equationr=LL−SbThe list price of the shirt is given as $30. Its sale price is shown to be $18. In order to find the discount rate r, we should simply substitute these values into the formula we found in the previous exercise.
r=LL−SS=18, L=30r=3030−18Subtract termr=3012ba=b/3a/3r=104Write as a decimalr=0.4
Written as a percent, the discount rate is 40%.|| |
| Exercises 34 aSolving the formula for m requires the isolation of m on one side.
d=VmLHS⋅V=RHS⋅VdV=mRearrange equationm=dVbThe density of the pyrite is given as 5.01 g/cm3 and its volume 1.2 g/cm3. In order to find the mass m, we substitute these values into the rewritten formula.
Therefore, the mass is 6.012 grams.|| |
| Exercises 35 To find the duration t, we need to solve the simple interest formula for t.
This new formula will help us determine how long we must leave the money in the account to earn $500 in interest. The annual interest rate is given as 4% which we express in decimal form as 0.04.
t=PrISubstitute valuest=2000⋅0.04500Multiplyt=80500ba=b/10a/10t=850Use a calculatort=6.25
We've found that t is 6.25 years. The decimal part can be converted to months by multiplying 12 by 41 and simplifying:
In other words, it takes 6 years and 3 months.|| |
| Exercises 36 To calculate the distance each flight is, we will need to use the distance formula.
Let's divide the above formula by r to isolate t.
Let d be the distance of each flight. The outbound flight averages 460 miles per hour and the return flight averages 500 miles per hour. Let's write an expression for the time taken in each flight.Outbound FlightReturn Flight
460d500dThe total flying time is the sum of the duration of both outbound and return flights. Moreover, we are also told that the total flying time is 4.8 hours.
Let's solve the above equation to find the value of d.
4.8=460d+500dba=b⋅500a⋅5004.8=460⋅500d⋅500+500dba=b⋅460a⋅4604.8=460⋅500d⋅500+500⋅460d⋅460Multiply4.8=230000500d+230000460dAdd fractions4.8=230000500d+460dAdd terms4.8=230000960dLHS⋅230000=RHS⋅2300004.8⋅230000=960dLHS/960=RHS/9609604.8⋅230000=dUse a calculator1150=dRearrange equationd=1150
Since d=1150 the distance of each flight is 1150 miles.|| |
| Exercises 37 aThe perimeter is made up of two straight sides and two semicircles at the ends. The two straight sides have each a length of x and the circular ends is each half the circumference of a circle with radius r. Together they make up a whole circle with radius r. Adding the pieces together we get:
P=x+x+πr+πr⇒P=2x+2πr.bWe need to isolate x on one of the sides.
P=2x+2πrLHS−2πr=RHS−2πrP−2πr=2xLHS/2=RHS/22P−πr=xRearrange equationx=2P−πrba=b1⋅ax=21P−πrcThe perimeter is given as 660 feet and the radius of the half circles is given as 50 feet. Inserting these values into the rewritten formula we can find x.
x=21P−πrP=660, r=50x=21⋅660−π⋅50b1⋅a=bax=2660−π⋅50Simplify quotient and productx=330−50πUse a calculatorx=172.9203673Round to nearest integerx≈173
When rounded to the nearest integer, the straight path x is 173 feet long.|| |
| Exercises 38 aThe distance d we travel in a car is expressed by two different equations. However, the traveled distance is the same in each case. So we can equate the given equations in order to relate g and t.
55t=20gbLet's solve this equation for g.
55t=20gLHS/20=RHS/202055t=gRearrange equationg=2055tba=b/5a/5g=411tcIf we travel 6 hours, then t=6. We can find the amount of gasoline the car uses to travel that distance using the equation found in part (b).
This means that we need 16 and a half gallons of gasoline for our trip. Furthermore, if we want to calculate how far we can travel, we need to use one of the equations given in the original problem. Let's use the second one.
We can travel 330 miles.|| |
| Exercises 39 aThe circumference of a circle is calculated using the formula:
where r is the radius. To calculate the radius, we can isolate it in the equation.
The formula for radius of a circle is then:
r=2πC.bTo calculate the radius of a column with a certain circumference, we have to substitute this circumference value for C in the formula
Let's do this for the desired circumferences.C2πCr
92π91.4cA cross section of this column will be a circle. We are asked how we can find the area of a circle when we know its circumference. The formula for the area of a circle is:
All we need to calculate the area is the radius, which we know how to find from the previous exercise.|| |
| Exercises 40 aWe have been told that the figure has two identical bases with equal side lengths b. Therefore, we can calculate the area of these two sides by squaring b and multiplying by 2:
The four other sides are also identical, each has a length l and a height b. Therefore, the combined area of these four sides are the product of l and b multiplied by 4:
To get the total surface area, we add the surface area of the base and the surface area of the sides:
A=2b2+4lb.bThe formula which we found in previous part of the exercise is:
Solving this for b could pose some problems as calculating the square root of an equation gives you both a positive and a negative solution. It's then much easier to leave b2 alone and just isolate l, if we are made to choose that is.|| |
| Exercises 41 The given temperatures are in different units. We need to convert one of the temperatures so that both are in the same unit. As the formulas for converting Fahrenheit to Celsius are equivalent, it does not matter which unit we choose to convert. We will arbitrarily choose to convert temperature of 20 degrees Celsius to Fahrenheit.
F=59C+32C=20∘F=59⋅20∘+32∘ca⋅b=ca⋅bF=5180∘+32∘Calculate quotientF=36∘+32∘Add termsF=68∘
We know that the Thermometer A displays 68∘ Fahrenheit. Thermometer B displays a temperature 70∘ Fahrenheit and that is greater than 68∘C. So your friend is wrong.|| |
| Exercises 42 Let's divide the figure into a triangle and a rectangle, labeling the base of the triangle a. Since the base of the figure is 8 cm and the base of the triangle is a, then the rectangle's length becomes 8−a.Now, let's write formulas for the area of the rectangle and the triangle:
Rectangle: Triangle: AR=h(8−a)AT=21ha.
The area of the whole figure AF is the sum of both areas. Let's add them and simplify the right-hand side.
AF=h(8−a)+21haFactor out hAF=h(8−a+21a)Simplify termsAF=h(8−2a)
We want to find a possible value of h and from the exercise we know that the area of the figure is 40 cm2. Now we can set up an equation that we solve for a.
solve for a
LHS/h=RHS/hh40=8−2aLHS−8=RHS−8h40−8=-2aChange signs-h40+8=2aLHS⋅2=RHS⋅2-h80+16=aRearrange terms16−h80=aRearrange equation
Now we are free to choose h, but it has to be a reasonable value that corresponds with the look of the figure. From the figure, we can tell that it should probably be less than the side that is 8 so let's try a few values that are less than 8 but larger than 0.h16−h80a
Of the lengths that we have calculated h=4 and 5 are not viable as they leave no room for the triangle's base a. From h=6 and 7, only 6 is a reasonable length as the triangle's base is obviously not more than half of the rectangle's length.|| |
| Exercises 43 Let's divide the regular polygon into 5 equal triangles, each having base b and height h.The area of the polygon is equal to the sum of the areas of the 5 identical triangles. We can calculate the area of a triangle using the formula:
By multiplying the above formula by 5 we get the formula for the area A of the regular polygon:
We want to solve this formula for the height h, so we will isolate it on one side.
A=25bhLHS⋅2=RHS⋅22A=5bhLHS/5b=RHS/5b5b2A=hRearrange equationh=5b2A|| |
| Exercises 44 Let's divide the regular polygon into 8 equal triangles, each having base b and height h.The area of the polygon is equal to the sum of the areas of the 8 identical triangles. We can calculate the area of a triangle using the formula:
By multiplying the above formula by 8, we get the formula for the area A of the regular polygon:
We want to solve this formula for height h, so we will isolate it on one of the sides.
A=4bhLHS/4b=RHS/4b4bA=hRearrange equationh=4bA|| |
| Exercises 45 We will solve the equation for a by isolating it on one side.
x=aba+b+cLHS⋅ab=RHS⋅ababx=a+b+cLHS−a=RHS−aabx−a=b+cFactor out aa(bx−1)=b+cLHS/(bx−1)=RHS/(bx−1)a=bx−1b+c|| |
| Exercises 46 We will solve the equation for a by isolating in on one side.
y=x(a−bab)Remove parenthesesy=x⋅a−baba⋅cb=ca⋅by=a−bxabLHS⋅(a−b)=RHS⋅(a−b)y(a−b)=xabDistribute yya−yb=xabLHS−xab=RHS−xabya−yb−xab=0LHS+yb=RHS+ybya−xab=ybFactor out aa(y−xb)=ybLHS/(y−xb)=RHS/(y−xb)a=y−xbyb|| |
| Exercises 47 According to the order of operations, exponents should be calculated first, followed by subtraction, then finally addition.
15−5+52Calculate power15−5+25Subtract term10+25Add terms35
You can use a calculator to work it out.|| |
| Exercises 48 According to the order of operations, exponents should be calculated first, followed by multiplications and divisions, then finally subtraction.
18⋅2−42÷8Calculate power18⋅2−16÷8Multiply36−16÷8Calculate quotient36−2Add terms34
You can use a calculator to work it out.|| |
| Exercises 49 According to the order of operations, exponents should be calculated first, followed by multiplications and divisions, then finally addition.
33+12÷3⋅5Calculate power27+12÷3⋅5Calculate quotient27+4⋅5Multiply27+20Add terms47
You can use a calculator to work it out.|| |
| Exercises 50 According to the order of operations, parentheses should be calculated first, followed by calculating exponents, then multiplications and divisions and finally addition.
25(5−6)+9÷3Subtract term25(-1)+9÷3Calculate power32(-1)+9÷3Multiply-32+9÷3Calculate quotient-32+3Add terms-29
You can use a calculator to work it out.|| |
| Exercises 51 || |
| Exercises 52 || |
| Exercises 53 || |
| Exercises 54 || |