We have two pieces of information:

1. A mobile phone and two calculators cost $110: $x + 2y = 110$ Where $x$ is the cost of one mobile phone and $y$ is the cost of one calculator.

2. Two mobile phones and three calculators cost $208: $2x + 3y = 208$

We can solve this system of equations to find the values of $x$ (cost of a mobile phone) and $y$ (cost of a calculator).

Step 1: Solve for one variable

We can solve the first equation for $x$ in terms of $y$: $x = 110 - 2y$

Step 2: Substitute into the second equation

Now substitute $x = 110 - 2y$ into the second equation: $2(110 - 2y) + 3y = 208$ Simplifying: $220 - 4y + 3y = 208$ $220 - y = 208$ $-y = 208 - 220$ $-y = -12$ $y = 12$

Step 3: Solve for $x$

Substitute $y = 12$ back into the first equation: $x + 2(12) = 110$ $x + 24 = 110$ $x = 110 - 24$ $x = 86$

Final Answer:

• The cost of one mobile phone is $86.
• The cost of one calculator is $12.

Would you like more details on how I solved this? Or have any other questions?

Here are 5 related questions:

1. How would you solve a system of equations using substitution or elimination?
2. What happens if there are more than two variables in a system of equations?
3. How can you check if your solution to a system of equations is correct?
4. What if we had three different equations for three unknowns, how would we solve that?
5. Can you solve this problem using matrices?

Tip: When solving systems of linear equations, substitution is a useful method when one equation is easy to solve for one variable.