Question

challenge: △ + ○ + ○ = 17
○ × △ + △ = 42

Explanation:

Step1: Let the triangle be $x$ and the circle be $y$.

Set up equations: $x + y + y=17$ (first - equation), $y\times x + x = 42$ (second - equation). From the first equation, we get $x=17 - 2y$.

Step2: Substitute $x = 17 - 2y$ into the second equation.

$(y\times(17 - 2y))+(17 - 2y)=42$. Expand it: $17y-2y^{2}+17 - 2y = 42$. Rearrange to get a quadratic equation: $-2y^{2}+15y - 25 = 0$. Multiply through by - 1 to get $2y^{2}-15y + 25 = 0$.

Step3: Solve the quadratic equation $2y^{2}-15y + 25 = 0$.

For a quadratic equation $ay^{2}+by + c = 0$ ($a = 2$, $b=-15$, $c = 25$), use the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-15)^{2}-4\times2\times25=225 - 200 = 25$. Then $y=\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}$. We get two solutions for $y$: $y_1=\frac{15 + 5}{4}=5$ and $y_2=\frac{15 - 5}{4}=\frac{5}{2}$.

Step4: Find the corresponding values of $x$.

If $y = 5$, then $x=17-2\times5=7$. If $y=\frac{5}{2}$, then $x=17-2\times\frac{5}{2}=12$. Check these values in the second - equation:

• When $x = 7$ and $y = 5$, $y\times x+x=5\times7 + 7=35 + 7=42$.
• When $x = 12$ and $y=\frac{5}{2}$, $y\times x+x=\frac{5}{2}\times12+12=30 + 12=42$.

Answer:

The triangle ($x$) is 7 and the circle ($y$) is 5.