Question

2x + y = 10x³
2x + 3y = 18 - 1

Explanation:

First, we simplify the equations. The first equation is \(2x + y = 10x^3\) (wait, maybe there's a typo? If it's \(2x + y = 10\) and the second is \(2x + 3y = 17\) (since \(18 - 1 = 17\)), we can use elimination. Subtract the first equation from the second: \((2x + 3y) - (2x + y) = 17 - 10\).

Step1: Simplify the second equation

The second equation \(2x + 3y = 18 - 1\) simplifies to \(2x + 3y = 17\).

Step2: Subtract the first equation from the second

Let the first equation be \(2x + y = 10\) (assuming the \(x^3\) is a typo). Then:
\[

$$\begin{align*} (2x + 3y) - (2x + y) &= 17 - 10 \\ 2x + 3y - 2x - y &= 7 \\ 2y &= 7 \\ y &= \frac{7}{2} = 3.5 \end{align*}$$

\]

Step3: Substitute y back into the first equation

Substitute \(y = 3.5\) into \(2x + y = 10\):
\[

$$\begin{align*} 2x + 3.5 &= 10 \\ 2x &= 10 - 3.5 \\ 2x &= 6.5 \\ x &= \frac{6.5}{2} = 3.25 \end{align*}$$

\]

If the first equation is indeed \(2x + y = 10x^3\), the problem becomes more complex (cubic equation), but the second equation is linear. Assuming a linear system (typo in \(x^3\)), the solution is as above.

Answer:

If the system is linear (correcting \(10x^3\) to \(10\)): \(x = 3.25\), \(y = 3.5\) (or \(x = \frac{13}{4}\), \(y = \frac{7}{2}\))