3. n(1) n(1) x 3 = n(2) [n(1) x 2] – 11 = n(3) n(1) + n(2) + n(3) = 64

Then guess and check

guess 100 so the first number is 100 the second number would be 300 and the third number would be 188 so obviously it is way less than 100. Keep guessing.

l=3w-4

2l+2w=64

then solve the second equation, substituting in 3w-4 for l

2(3w-4)+2w=64

6w-8+2w=64

8w-8=64

8w=56

w=7

then solve the first equation, substituting 7 for w

l=3(7)-4

l=21-4

l=17

so the length is 17 and the width is 7

2. In a triangle, the sum of the angles is always 180 degrees, so start there. Assign the first angle the variable “a”, the second “b” and the third “c”

a+b+c=180

if the second angle measures twice the first, you can say

b=2a

then if the third angle measures 5 more than the second, you can say

c=b+5

then substitute 2a for b

c=2a+5

a+2a+2a+5=180

first solve for a

5a+5=180

5a=175

a=35

The second angle is twice the first, so the second angle is 70

The third angle is 5 more than the second, so the third angle is 75

the first angle is 35, the second is 70, and the third is 75. These add up to 180.

3. This problem is a lot like the last one, assign the numbers the variables a, b, and c.

a+b+c=64

the second number is 3 more than the first, so

b=3+a

the third number is 11 less than twice the first, so

c=2a-11

so then with substitution

a+b+c becomes a+3+a+2a-11=64

then solve

a+3+a+2a-11=64

4a-8=64

4a=72

a=18

so if a=18, then b=18+3, or 21, and c=2(18)-11 or 25

2(l + w) = 64

2l + 2w = 64

2(3w – 4) + 2w = 64

6w – 8 + 2w = 64

8w = 72

w = 9

l = 23

Ans. width = 9cm, length = 23cm

2) b = 2a

c = 2b + 5 = 2(2a) + 5 = 4a + 5

a + b + c = 180

a + 2a + 4a + 5 = 180

7a = 175

a = 25

b = 2(25) = 50

c = 180 – 75 = 105

Ans. Angles are 25, 50, and 105 degrees

3) a + b + c = 64

b = a + 3

c = 2a – 11

a + (a + 3) + (2a – 11) = 64

4a – 8 = 64

4a = 56

a = 14

b = 17

c = 23

P = 2L+2W

64 = 2(3W-4)+2W

64 = 6W-8+2W

64 = 8W-8

W=72/8

W = 9

L = 27-4

L = 23

2.

180 = x+2x +(2x+5) each term being the angles resp.

180 = 5x+5

180-5 = 5x

5x = 175

x = 35° first angle

2x = 70° second angle

2x+5 = 75° third angle

3.

64 = x+(x+3)+(2x-11) each term being the numbers resp.

64 = 4x-8

4x = 72

x = 18 first number

x+3 = 21 second number

2x-11 = 25 third number

P = 2L + 2W = 64cm

Now, use the 3W – 4 expression for L in the second equation.

P = 2(3w-4) +2w = 64CM

Solve for W, then plug THAT answer into the first equation to solve for L.

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The trick to problem 2 is to remember that the sum of the angles of a triangle = 180 deg. Lets name the angles A, B and C. We know A + B + C = 180.

2A = B so A = 1/2B

B+5 = C

Now plug in those expressions for A, and C in our first equation and solve for B. Then plug B into the other two equations to solve for A and C. You can do that.

—————–

3. X + Y + Z = 64 is the first given.

X+3 = y is the second.

2x – 11 = Z is the third.

Plug in the values for Y and Z into the first equation and solve for X. Then plug THAT number into the other two and solve for Y and Z.

L=3W-4

64 = 2W + 2(3W-4)

64 = 2W + 6W -8

64= 8W-8

72=8W

W=9

L=3W-4

L=3(9)-4

L=23

2) 180= A1 + A2+A3

A2 = 2*A1

A3=A2+5

A3=2*A1 +5

180 = A1 +(2*A1) + (2*A1+5)

A1=x

180 = x+ 2x + 2x + 5

180 = 5x +5

175 = 5x

x=A1=35

A2=2*A1 = 2(35) = 70

A3 = A2+5 = 70+5 = 75

3) x + y+ z=64

y = x+3

z=2x-11

x+(x+3) +(2x-11)=64

x+ x+3+2x-11=64

4x-8=64

4x=72

x=18

y=x+3 = 18+3=21

z=2x-11 + 2(18) -11 = 36-11=25

the third has i typo i think