Essential Mathematics 1: Algebra and Trigonometry Assignment One
Question 1a)
Solve 2+x=3x+2x-3. Leave solutions in simplest rational form. The linear equation which is in the form ax+b=0 or can be transformed into an equivalent equation into this form.

2+x=3x+2x-3 Expand 2x-3. 2+x=3x+2x-6 Add 3x+2x together. 2+x=5x-6 Subtract 5x from both sides. 2-4x=-6 Subract 2 from both sides -4x=-8 Divide both sides by-4 x=2
Check Solution x=2
2+x=3x+2x-3 Change x to 2
2+2=3×2+22-3 Add 2+2 together, multiply 3 and 2, expand 22-3 4=6+4-6 Subtract 6 from 6+4 4=4

Thus, the solution for 2+x=3x+2x-3 is x=2 .

Question 1b)
Solve 2xx-1+3x=x-9xx-1 This equation is rational, it can be written as the quotient of two polynomials. In addiction of expressions with unequal denominators, the result is written in lowest terms and expressions are built to higher terms using the lowest common denominator.

2xx-1+3x=x-9xx-1 multiply x-1 from 3x

2+3x-1=x-9 Expand 3x-1 2+3x-3=x-9 Subtract 3 from 2 3x-1=x-9 Subtract x from both sides 2x-1=-9 Add 1 from both sides 2x=-8 Divide 2 from both sides x=-4

Check Solution x=-4 2+3x-1=x-9 Change x to-4
2+3-4-1=-4-9 Subtract 4 from-1, Subtract 4 from-9 2+3-5=-13 Expand 3(-5) 2-15=-13 Add 2 to-15 -13=-13 Both sides are equal to-13
The solution to this equation 2xx-1+3x=x-9xx-1 is x=-4.
Question 1c)

Solve -2<3x+2+1<10, this equation is linear, combined inequality can be solved by isolating the variable in the middle.

-2<3x+2+1<10 Expand 3(x+2)
-2<3x+6+1<10 Add 6 and 1 together
-2<3x+7<10 Subtract 7 by all sides
-9<3x<3 Divide 3 by all sides
-3<x<1
(-3,1)

The solution set for this equation -2<3x+2+1<10 is (-3,1).

Question 1d)

Solve x2-x-12≥0, this inequality equation which is on the left side can be written as a quotient linear factor that can be solved through a sign diagram.

x2-x-12≥0 Factorise the left side x-4x+3≥0 x=4 x=-3

The critical points are 4 and -2 where respectively, x-4 and x+3 are zero. Draw a number line showing the critical points

x-4 | - | - | + | x+3 | - | + | + | (x-4)(x+3)≥0 | + | - | + |

-3 4 x

The critical points divide the real number line into the intervals -∞, -3, -3,4and 4, ∞.
The product has a sign, respectively, positive, negative and positive in these intervals. The equation part of the inequality is satisfied at both critical points and the inequality holds when (x-4)(x+3)≥0 is positive, hence the solution set is -∞,-3 ∪4, ∞.

Question 2a)

Simply 3x2y-3z-5-2 x4 and write without negative powers, this is an exponential expression and can be simplified when no unnecessary parentheses appear, no powers raised to powers, no negative or zero exponents appear and each base only occurs once.

3x2y-3z-5-2 x4 Multiply the power of -2 to every exponent inside the brackets

-9x-4y6z10 x4 Remove the brackets

-9x-4y6z10x4 Add x4and x-4 together

-9y6z10 Change the exponents into positive exponents

y6z109

The solution is y6z109.

Question 2b)

Simply (2x4y5zxy-5z2) -1 and write without negative powers, this is an exponential expression and can be simplified when no unnecessary parentheses appear, no powers raised to powers, no negative or zero exponents appear and each base only occurs once.

(2x4y5zxy-5z2) -1 Multiply the power of -1 to every exponent inside the brackets

-2x-4y-5z-1x-1y5z-2 Change the exponents into positive exponents

z z2x3y10

The solution is z z2x3y10.

Question 2c)

Compute 33-2 rationalising the dominator, leave your answer in exact form with surds, no decimals. A convention of writing a rational number in simplest form is to avoid having radicals in the denominator. We can build up the fraction to find an equivalent expression.

33-2 Build up using 3+2 as the numerator and denominator

33-2 x 3+23+2 Multiply the numerator and the denominator by the conjugate expression for the denominator

3(3+2)(3-2)(3+2) Factorise numerator and the denominator

9+329+32-32-2 Subtract 32 from 32

9+329-2 Subtract 2 from 9

9+327

The solution is 9+327.

Question 3a)

Rearrange the equation PV=nRT for T. In these situations, letters are used as coefficients rather than particular numbers. However, the procedures, for solving for a specified variable are essentially the same; the other variables are simply treated as constants. The equation is linear in T, the specified variable.

PV=nRT for T Divide nR both sides

PVnR=nRTnR Swap both sides, putting the T to the left

T= PVnR

The solution is T= PVnR.

Question 3b)

Rearrange the equation 1x=1y+1z for z, this equation is linear in z, the specified variable.

1x=1y+1z for z Multiply the denominator to each fraction

xyz ×1x=xyz ×1y+xyz ×1z Remove common variables for each fraction

yz-x-z=xz-x-z+xy Factorize yz-x-z

-zx-y=xy Divide x-y both sides

-z(x-y)=xy(x-y) Make z positive by multiplying the negative sign to both sides

z=-xy(x-y)

The solution is z=-xy(x-y).

Question 3c)

Rearrange the equation T=C1-r for r. This equation is linear in r, the specified variable.

T=C1-r for r. Multiply 1-r both sides

1-r × T=C1-r × 1-r Divide T both sides

1-r × TT=CT Subtract 1 to both sides

-r=-1+CT Make r positive by multiplying the negative sign to both sides

r=1-CT

The solution is r=1-CT.
Question 4a)

Solve 3x2+6x+2=0, this is an quadratic equation which is in form of ax2+bx+c=0, where a,b, and c are real numbers and a≠0 and can be solved by using the quadratic formula x=-b±b2-4ac2a, leave answers in simplest surd form.

3x2+6x+2=0 Use quadratic formula

x=-b±b2-4ac2a Use quadratic equation to find a, b and c

a=3 b=6 c=2 Put a=3,b=6 and c=2 into quadratic formula

x=-6±62-4×3×22×3 Square 6, multiply 4×3×2, and multiply (2×3)

x=-6±36-246 Subtract 24 from 36

x=-6±126 Put 12 into simplest radical form

x=-6±236 Put equation into simplest surd form by dividing it by 2

x=-3±33 Separate ± into two answers

x=-3+33 or x=-3-33

The solutions are x=-3+33 or x=-3-33.

Question 4b)

Use the discriminant b2-4ac to show that the equation x2-4x+5=0 does not have a real solution. Then find the complex solutions using i=-1

x2-4x+5=0 Use quadratic equation to find a, b and c

a=1 b=-4 c=5 Put a=1,b=-4 and c=2 into discriminant b2-4ac

x=(-4)2-4×(1)×(5) Square -4, and multiply 4×(1)×(5)

x=16-20 Subtract 20 from 16

x=-4 Zero real solutions

∆<0 Use quadratic formula

x=-b±b2-4ac2a Put a=1,b=-4 and c=2 into quadratic formula

x=-(-4)±(-4)2-4×(1)×(5)2×1 Multiply – to -4, square -4, multiply 4×1×5 and 2×1

x=4±16-202 Subtract 20 from 16

x=4±-42 Change -4 into a complex number

x=4±4(-1)2 (-1) becomes an imaginary number (i)

x=4±4i2 Square root 4i

x=4±2i2 Divide the nominator by 2

x=2±i Separate ± into two answers

x=2-i or x=2+i

The solutions are x=2-i or x=2+i .

Question 5a)

Solve the equation 2xx-3+(x-3)2=0 by factoring, do not expand first. We can remove the nonmonomial common factors to solve the equation.

2xx-3+(x-3)2=0 Remove the common factors x-32x+x-3=0 Simplify 2x+x-3 x-33x-3=0 Factorise [3x-3] x-3 3x-1=0 In its lowest terms x-3(3x-3)=0 Expand x-3(3x-3
3x2-3x-9x+9=0 Subtract 3 from -9
3x2-12x+9=0 Find the answers that equals to zero (zero law) x=3 or x=1

x=3 Check x=3 by substituting back
3x2-12x+9=0 Change x into 3
3×(3)2-12×(3)+9=0 Multiply 3×(3)2and 12×(3)
27-36+9=0 Subtract 36 from 27
-9+9=0 Subtract 9 from 9
0=0

x=1 Check x=1 by substituting back
3x2-12x+9=0 Change x into 1
3×(1)2-12×(1)+9=0 Multiply 3×(1)2and 12×(1)
3-12+9=0 Subtract 12 from 12
0=0

The solutions are x=3 or x=1.

Question 5b)

Solve the equation 3u+42-53u+4-14=0 by factoring, do not expand first. We can remove the nonmonomial common factors to solve the equation.

3u+42-53u+4-14=0 Remove the common factors
3u+43u+4-5-14=0 Expand 3u+4-5
3u+43u-1-14=0 Add14 from both sides
3u+43u-1=14 Expand 3u+43u-1
9u2-3u+12u-4=14 Subtract 3u from 12u
9u2+9u-4=14 Find the answers that equals to 14 u=1 or u=-2

u=1 Check u=1 by substituting back
9u2+9u-4=14 Change u into 1
9(1)2+9(1)-4=14 Multiply 9×(1)2 and 9×(1)
9+9-4=14 Add 9 and 9 together
18-4=14 Subtract 4 from 18
14=14

u=-2 Check u=-2 by substituting back
9u2+9u-4=14 Change u into -2
9(-2)2+9(-2)-4=14 Multiply 9×(-2)2 and 9×(-2)
36-18-4=14 Subtract 4 from -18
36-22=14 Subtract 22 from 36
14=14

The solutions are u=1 or x=-2.
Question 5c)

Solve t4-2t2-8=0 for real values of t. Making a substitution that transforms it into a quadratic can solve this. Introducing the substitution of u=t2 then t4=(t2)2 and the equation becomes:

t4-2t2-8=0 Substitution of u=t2 u2-2u-8=0 Factor the integers u-4u+2=0 Zero law of u-4=0 and u+2=0 u=4 or u=-2

u=4 Undo the original substitution u=t2 t2=4 Square root both sides t=±4 Separate ± into two answers t=-2 or t=2

u=-2 Undo the original substitution u=t2 t2=-2 Square root both sides t=±-2 Change -2 into a complex number t=±2i Separate ± into two answers t=2i or t=-2i

The solutions are t=-2 or t=2 or t=2i or t=-2i

Question 6) Consider the complex numbers w=4+2i and z=2-i. Find in standard form for complex numbers.

a) w+z Change w to 4+2i and z to 2-i
4+2i+2-i Add 4 and 2 together, subtract I from 2i
6+i

The solution is 6+i

b) w-z Change w to 4+2i and z to 2-i
4+2i-2-i Multiply – to 2-i
4+i-2+i Subtract 2 from 4, add 2i and i together
2+3i

The solution is 2+3i

c) wz Change w to 4+2i and z to 2-i
4+2i2-i Expand 4+2i2-i
8-4i+4i-2i2 Subtitute i2 =-1 into-2i2
8-4i+4i+2 Subtract -4i from 4i, add 8 and 2 together wz=10 The solution is 10

d) wz Change w to 4+2i and z to 2-i

4+2i 2-i Multiply the numerator and the denominator by the conjugate expression for the denominator

4+2i 2-i×2+i 2+i Factor the nominators and denominators

(4+2i)(2+i) (2-i)(2+i) Expand 4+2i2+i and (2-i)(2+i)

8+4i+4i+2i2 4+2i-2i-i2 Add 4i and 4i together and subtract 2i from 2i

8+8i+2i2 4-i2 Subtitute i2 =-1 into+2i2 in nominator and-i2 in the denominator

8+8i-2 4+1 Subtract 2 from 8 and add 5 and 1 together

6+8i 5 or 65+8i5

The solutions are 6+8i 5 or 65+8i5.