Solve the first equation for one variable (for example, $y = 7 - x$) and substitute into the second equation, or think of $x$ and $y$ as the roots of a quadratic with sum 7 and product 10.

Once you get a quadratic equation, you will find two possible values. Both will be candidates for $x$ and $y$. Use the condition $x \ge y$ to decide which one is $x$.

In Desmos, type y = x^2 - 7x + 10 to graph the quadratic equation that has $x$ and $y$ as its roots.

Look at where the graph crosses the x-axis (the x-intercepts). These x-values are the two possible numbers that satisfy the system. Use the condition $x \ge y$ to choose the larger of these values as $x$.

If two numbers $x$ and $y$ have sum $x + y = 7$ and product $xy = 10$, then they can be seen as the two roots of a quadratic equation.

For roots $x$ and $y$:

• Their sum is $x + y$, which equals the coefficient of $t$ with a negative sign.
• Their product is $xy$, which is the constant term.

So $x$ and $y$ are the roots of the equation $t^2 - (x + y)t + xy = 0$.

You know $x + y = 7$ and $xy = 10$, so substitute these into the quadratic form:

becomes

Now you just need to solve this quadratic equation for $t$.

Factor the quadratic $t^2 - 7t + 10$.

Look for two numbers that multiply to $10$ and add to $7$. Those numbers are $5$ and $2$, so

Set each factor equal to zero:

• $t - 5 = 0$ gives $t = 5$.
• $t - 2 = 0$ gives $t = 2$.

So the two possible values for $x$ and $y$ are $5$ and $2$ (in some order).

You are told that $x \ge y$, meaning $x$ is the larger (or equal) number.

From the previous step, the two numbers are $5$ and $2$, and the larger one is $5$.

Therefore, $x = 5$.