Maths Olympiad Prep

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Algebra Difficulty 1.0 Junior Prove it Canada

IMG0 What is the integer tt for which 2t3+3t2=26\dfrac{2t}{3} + \dfrac{3t}{2} = 26?
Figure 1 What is the integer xx for which 3+x4=6+x8\dfrac{3+x}{4} = \dfrac{6+x}{8} ?
Figure 2 Suppose that y>0y > 0 and 32+42+122=32+42+y2\sqrt{3^2+4^2+12^2} = \sqrt{3^2+4^2} + \sqrt{y^2}. Determine the value of yy.

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Solution

Multiplying both sides of the equation by 23=62 \cdot 3 = 6, we obtain 62t3+63t2=6266 \cdot \dfrac{2t}{3} + 6\cdot\dfrac{3t}{2} = 6 \cdot 26 or 4t+9t=1564t + 9t = 156.

Simplifying, we obtain 13t=15613t = 156 and so t=12t = 12. Alternatively, using a common denominator of 23=62 \cdot 3 = 6, we obtain 22t23+33t32=26\dfrac{2 \cdot 2t}{2 \cdot 3} + \dfrac{3 \cdot 3t}{3 \cdot 2} = 26 or 4t6+9t6=26\dfrac{4t}{6} + \dfrac{9t}{6} = 26. Simplifying, we obtain $13t6\$\dfrac{13t}{6} =
26andso and so 13t = 6 \cdot 26or or t = 6 \cdot 2 = 12. Multiplying both sides of the equation by

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Figure for this problem8,weobtain, we obtain 8(3+x)4\dfrac{8(3+x)}{4} = 6 + xwhichsimplifiesto which simplifies to 2(3+x) = 6 + x.Simplifyingfurther,weobtain. Simplifying further, we obtain 6 + 2x = 6 +
xandso and so x = 0. Alternatively, splitting each fraction into two pieces, we obtain

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Figure for this problem

Figure for this problem34+x4=68+x8\dfrac{3}{4} + \dfrac{x}{4} = \dfrac{6}{8} + \dfrac{x}{8}.Since. Since 34=68\dfrac{3}{4} = \dfrac{6}{8},weobtain, we obtain x4=x8\dfrac{x}{4} = \dfrac{x}{8}andso and so 8x = 4xor or x = 0.Since. Since y > 0,then, then y2\sqrt{y^2} = y. From the given equation, we obtain

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Figure for this problem

Figure for this problem9\sqrt{9} +
16 + 144} = 9\sqrt{9} + 16} + y.Thus,. Thus, 169=25\sqrt{169} = \sqrt{25} + yor or 13 = 5 + yandso and so y = 8$.

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