Part a) There are two clusters of shapes. Both clusters are equal to eachother. The first cluster has five circles and three triangles. The second cluster has three circles and six triangles. The value of the triangle is 9. What is the value of the circle?
Part b) There are two clusters that are equal to eachother. The first cluster has two squares, two triangles and a circle. The second cluster has four squares, two triangles and two circles. Each cluster has an overall value of 6. Find the value of the triangle if the value of the circle is twice that of the square.
a)
5x + 27 = 3x + 54
2x = 27
x = 13.5
b)
2s + 2t + c = 6
4s + 2t + 2c = 6
c = 2s
then solve by substitution
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Part a)
5C + 3T = 3C + 6T
2C = 3T
and we know that T = 9
so 2C = 27
C = 27/2 = 13.5
check: 5C + 3T = 5(27/2) + 3(9) = 67.5 + 27 = 94.5
3C + 6T = 3(27/2) + 6(9) = 40.5 + 54 = 94.5 yep
Part b:
2S + 2T + C = 6
4S + 2T + 2C = 6
subtracting the first from the second:
2S + C = 0
and we know that C = 2S
subtituting: 2S + 2S = 0
4S = 0
S = 0
C = 2S = 0
since S = C = 0, that leave 2T = 6 for both clusters
T = 3
so S = C = 0 and T = 3 satisfies the conditions
References :
a)
5x + 27 = 3x + 54
2x = 27
x = 13.5
b)
2s + 2t + c = 6
4s + 2t + 2c = 6
c = 2s
then solve by substitution
References :
the easiest way to analyze these problems are by setting up pictures (which i can’t show you) or making the sentences equations: here we go
part a)
two equal clusters-
5 circles + 3 triangles = 3 circles + 6 triangles
5C + 3T = 3C + 6T with T=9
5C + 3*9 = 3C + 6*9
5c + 27 = 3C + 54 subtract 3C from both sides
2C + 27 = 54 subtract 27 from both sides
2C = 27
C = circles = 27/2 = 13.5
part b)
2 squares + 2 traingles + 1 circle = 4 squares + 2 triangles + 2 circles
2s + 2t + 1c = 4s + 2t + 2c = 6 and c=2s
substitute 2s for every c gives
2s + 2t + 2s = 4s + 2t + 2*2s
4s + 2t = 8s + 2t both sides of this equation = 6 and we can set the equation = to each other–>
4s + 2t = 8s + 2t subtract 2t from both sides and we get
4s = 8s which is impossible except for s=0
so c=2s–> c=0 sustituting into the original equations..
2s + 2t + 1c = 4s + 2t + 2c = 6 gives us
2*0 + 2t +0 = 4*0 + 2t +2*0 = 6
now we have 2t = 6 so…
t=3
i hope that part b is correct, i think it is, and i hope this helps
References :