Question Video: Solving Systems of Linear and Quadratic Equations | Nagwa Question Video: Solving Systems of Linear and Quadratic Equations | Nagwa

Question Video: Solving Systems of Linear and Quadratic Equations Mathematics • Third Year of Preparatory School

Given that 𝑥² + 𝑥𝑦 = 18, and 𝑥 + 𝑦 = 6, find the value of 𝑥.

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Video Transcript

Given that 𝑥 squared plus 𝑥𝑦 equals 18 and 𝑥 plus 𝑦 equals six, find the value of 𝑥.

What we have here is a linear-quadratic system of equations. The first equation 𝑥 plus 𝑦 equals six is linear. And the second equation 𝑥 squared plus 𝑥𝑦 equals 18 is quadratic because it includes an 𝑥 squared term and also the term 𝑥𝑦, where the two variables are multiplied together. We’re not asked to fully solve this system of equations but simply to find the value of 𝑥. So we’re going to do this using the substitution method.

We begin by rearranging the first equation to give 𝑦 equals six minus 𝑥 because this gives an expression for 𝑦 in terms of 𝑥, which we can substitute into the second equation to give an equation in 𝑥 only. Doing so gives the equation 𝑥 squared plus 𝑥 multiplied by six minus 𝑥 equals 18. And we now have a quadratic equation in 𝑥, which we can solve.

We distribute the parentheses on the left-hand side to give 𝑥 squared plus six 𝑥 minus 𝑥 squared equals 18. And we now see that the 𝑥 squared and negative 𝑥 squared terms will cancel each other out. So, in fact, our equation reduces to a linear equation. We have six 𝑥 is equal to 18. And this equation can be solved by dividing each side by six to give 𝑥 equals three. So by substituting 𝑦 equals six minus 𝑥 into the second equation, we created an equation in 𝑥 only which we could then solve to find the value of 𝑥.

Of course, we aren’t asked to find the value of 𝑦 in this problem. But if we did need to, we could substitute the value of 𝑥 we’ve just found back into our linear equation, 𝑦 equals six minus 𝑥, to find the corresponding value of 𝑦. Our solution to the problem is that the value of 𝑥 is three.

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