Equation Solving Techniques

Lesson · Intermediate

Algebra: Equations

\text{Similar Equations}

\text{Substitution}

2022 Philippine Mathematical Olympiad — Qualifying Stage

Let $a,b,c$ be real numbers such that

3ab+2=6b, \qquad 3bc+2=5c, \qquad 3ca+2=4a.

Suppose the only possible values for the product $abc$ are $r/s$ and $t/u$ , where $r/s$ and $t/u$ are both fractions in lowest terms. Find $r+s+t+u$ .

3ca+2=4a \quad\Rightarrow\quad 3ca=4a-2 \quad\Rightarrow\quad c=\frac{4a-2}{3a}.

3ab+2=6b \quad\Rightarrow\quad 2=b(6-3a) \quad\Rightarrow\quad b=\frac{2}{6-3a}.

Substitute these expressions for $b$ and $c$ into $3bc+2=5c$ :

3 \left( \frac{2}{6-3a} \right) \left( \frac{4a-2}{3a} \right) +2 = 5 \left( \frac{4a-2}{3a} \right).

2(4a-2)+2(2-a)(3a) = 5(4a-2)(2-a).

8a-4+12a-6a^2 = 40a-20a^2-20+10a.

14a^2-30a+16=0.

7a^2-15a+8=0.

(7a-8)(a-1)=0.

\Rightarrow\quad a=1 \qquad\text{or}\qquad a=\frac87.

Case 1: $a=1$

b = \frac{2}{6-3a} = \frac23,

c = \frac{4a-2}{3a} = \frac23.

abc = 1\cdot\frac23\cdot\frac23 = \frac49.

Case 2: $a=\frac87$

b = \frac{2}{6-3a} = \frac{2}{6-\frac{24}{7}} = \frac79.

c = \frac{4a-2}{3a} = \frac{ \frac{32}{7}-2 }{ \frac{24}{7} } = \frac34.

abc = \frac87\cdot\frac79\cdot\frac34 = \frac23.

\frac rs=\frac49, \qquad \frac tu=\frac23.

\boxed{ r+s+t+u = 4+9+2+3 = 18 }

Back to Toolkit