Question for courseworkhero.co.uk

 This is for  courseworkhero.co.uk. I made some comments about some of the answers you had 

Can you answer these few questions that I had. I wrote them in the document (attached—scroll down)

I appreciate it if you can.

 

Thanks

Running Head: ELEMENTARY ALGEBRA

ELEMENTARY ALGEBRA 9

Elementary Algebra

Name

Institution Affiliation

Elementary Algebra

Question 2

x
=
3/2y +2 (Equation 1)

2x -3y
=
9
(Equation 2)

Substituting x in equation 2;

2(3/2y +2) -3y
= 9

3y +4-3y
= 9

4
=9

Therefore, because the equation has no solution, then it can be said that: (e)no solution, inconsistent system.

Question 3

1/x + 1/y
=
5/6 (Equation 1)

1/x-
1/y
=
1/6 (Equation 2)

Solution

Substituting a=1/x and b=1/y then the linear equation will be

a + b
=
5/6

a –b
=
1/6

Multiplying every element with the denominator, which is 6, we will have and add the two linear equations,

6a + 6b
=
5

6a- 6b
=
1

12a
=
6

12a/12
=
6

a =
½

Substituting a
=1/2 using the first equation then

6(1/2) + 6b
=
5

3 +
6b
=
5

6b
=
5-3

6b
=
2

b =
1/3

Since a =1/x the value of x will be ½ = 1/x then multiplying both sides 2x as a common factor;

½(2x)
=
1/x (2x)

x
=
2

Also since b= 1/y, 1/3
= 1/y, then multiplying both sides by 3y as a common factor then,

1/3 (3y)
=
1/y (3y)

Y
=
3

The value of x and y are 2 and 3 respectively.

Solution c. (2, 3)

Question 5

4x + 2 y +3z
=
6 (Equation 1)

2x – 3y – 4z
=
-4 (Equation 2)

8x + 4y + 6z
=
12 (Equation 3)

Making x the subject of the equation 1, then x= 6-2y/4 -3z/4 which becomes x = 3/2-1/2y- 3z/4.

Substitution for x in equation 2 then

2(3/2-1y/2-3z/4) -3y -4z
=
-4

3-y -3z/2-3y -4z
=
-4

3-4y – 11z/2
=
-4

Multiplying every side by 2 which is the common factor, then

8y +11z
=
14 (Equation 3)

Substitution for x in equation 3 then

8(3/2-1/2y- 3z/4) + 4y + 6z
=
12

4y –
6z + 4y + 6z
=
12

8y
=
12

y
=
12/8

y
=
3/2

The answer is b: The system is inconsistent as there are infinite solutions

Question 8

[(3a2b3z-5) / 2a-3b-2z5)]-3

Solution

Inversing the indices changes the power of -3 to 3.

= [(2a-3b-2z5)/ (3a2b3z-5)] 3

= (8a-9b-6z15)/ (27a6b9z15) (following the law of indices on division, the powers are subtracted.

= 8z30/27a15b15 (the answer is a).

Question 14

4+3i/4+i

Solution

= (4+3i/4+i) this is then multiplied by the conjugate, which is (4-i)

= (4+3i/4+i)/(4-i)/(4-i)

= (16-4i+12i-3i2)/(16-4i+4i-i2)

= 16-8i-3(-1)/16-(-1) (in this case i2=-1)

= 19-8i/17

= 19/17-8i/17 (the answer is e)

Question 4

(x+2)/(x-2)- (x-6)/2-x)

Solution

The common factor in this equation is (x-2)(2-x)

(x+2)/(x-2)- (x-6)/2-x)
= ((x+2) (2-x)-((x-6)(x-2))/(x-2)(2-x)

= (2x-x2+4-2x)-(x2-2x-6x+12)/ (x-2)(2-x)

= (4- x2)-( x2-8x+12)/ (x-2)(2-x) can you explain how did you get to this step from the one before= (4-2×2+8x+12)/ (x-2)(2-x)

= -2×2+8x-8 (then divide by 2 becomes -x2+4x-4

= -(x2 -4x+4)/ (x-2)(2-x)

= -(x22x-2x+4)/ (x-2)(2-x)

= -(x(x-2)-2(x-2) (x-2) (2-x)

=-(x-2) (x-2)/ (x-2)(2-x)

=-(x-2)/(2-x)

= -(x-2) /-(x-2)

= -1

Question 5

2x/(x2+11x+18)+2x/x2-4

Solution

Common factor of the equation is (x+9)(x+2)(x-2)

= 2x(x-2) +2x(x+9)/ (x+9) (x+2)(x-2)

= (2×2-4x+2×2+18x)/ (x+9) (x+2)(x-2)

= (4×2+14x)/ (x+9) (x+2)(x-2)

= 2x (2x+7)/ (x+9) (x+2) (x-2), therefore the answer is a.

Question 7

X3+1 divided by x-1

Solution

From identities x3+1 is (x2-x+1) (x+1) thus dividing with x-1;

This will become x2-x+1+2/x-1, thus the answer to this is a.

Question 8

((-10b)2)1/2

Solution

= (-10b)2

= 100b2

= (100b2)1/2

= 10/b/, thus the answer is b.

Question 3

Multiply [(6a +b) +4]2

Solution

= (6a+b) (6a +b) +16)

= 36a2+6ab +6ab+ b2 +42

= 36a2 +12ab+ 16

Question 4

Factor (x-y) 2 + 3(x-y)-10

Let take (x-y) to be m, therefore, the equation will be;

= m2 + 3m-10

= m2 +5m -2m-10

= (m+5) –2(m+5)

= (m+5) (m-2)

Then substituting m for (x-y) the equation will be (x-y+5) (x-y-2)

Question one

Simplify the rational expression

T3 -5t2 +6t/9t-t3

Solution

= t (t2-5t +6)/t (9-t2), from polynomials 9-t2 is – (t-3) (t+3)

= t (t-2) (t-3)/-t (t-3) (t+3) where did the first T in the answer. Wouldn’t the answer be t(t-2)(t+3)

= – (t-2) (t-3)/ (t-3) (t+3)

= – (t-2)/ (t+3)

Question 2

Divide the rational expressions Can you write this out on paper and show me because its difficult to see how you came to your answer

(2×2+5xy+3y2/ 3×2-5xy +2y2) ÷ (2×2+xy-3y2)/3×2-5xy+2y2)

Simplifying the individual quadratic equations;

= (2×2 +2xy +3xy +3y2/3×2-3xy -2xy +2y2) ÷ (2×2 +3xy -2xy -3y2/3×2-3xy-2xy+2y)

= (2x+y) (2x+3y)/ (3x-2y) (x-y) y÷(x-) (2x +3y)/ (3x-2y) (x-y)

= (2x+y) (2x+3y)/ (3x-2y) (x-y) x ((3x-2y) (x-y) (x-) (2x +3y) (this entails reversing the sign to multiplication, which means the second equation after the division sign is inverted. And after cancelling the like terms, the final answer will be:

= (2x+y)/x-y