Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns | Nagwa Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns | Nagwa

# Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns Mathematics • Third Year of Preparatory School

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A right triangle has a hypotenuse of 35 cm and a perimeter of 84 cm. Find the lengths of the other two sides.

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### Video Transcript

A right triangle has a hypotenuse of 35 centimeters and a perimeter of 84 centimeters. Find the lengths of the other two sides.

First, let’s sketch what we know. We know that the hypotenuse is the side opposite the right angle and that it measures 35 centimeters. The two other sides are unknowns, and that is what we’re trying to find. Additionally, we’ve been given that the perimeter of this right triangle is 84 centimeters. To solve this problem, we’ll need to form some equations. Since the perimeter is the distance all the way around the triangle, we can form an equation that says 𝑥 plus 𝑦 plus 35 equals 84. We can isolate the variables on the left by subtracting 35 from both sides of this equation to get 𝑥 plus 𝑦 equals 49. We’ll let this be our first equation.

To create a second equation, we can use the Pythagorean theorem since we’re working with a right triangle. The Pythagorean theorem 𝑎 squared plus 𝑏 squared equals 𝑐 squared tells us that the sum of the squares of the two smaller sides in a right triangle will be equal to the square of the hypotenuse. Substituting what we know about our triangle, we can come up with an equation that is 𝑥 squared plus 𝑦 squared equals 35 squared. By squaring 35, we find 𝑥 squared plus 𝑦 squared equals 1225, which becomes our second equation.

We can now use strategies to solve simultaneous equations to find values for 𝑥 and 𝑦. We could solve this by rewriting the first equation where one variable is in terms of the other and substituting that into our second equation. We can make 𝑥 the subject of our first equation by subtracting 𝑦 from both sides so that we have a third equation that tells us 𝑥 equals 49 minus 𝑦. Equation three is equivalent to equation one but just written in a different form. We’ll take equation three and substitute it in for the 𝑥-value in equation two, which gives us 49 minus 𝑦 all squared plus 𝑦 squared equals 1225.

In order to solve for 𝑦, our first step is to expand the term 49 minus 𝑦 squared. We know that the form 𝑎 minus 𝑏 squared will equal 𝑎 squared minus two 𝑎𝑏 plus 𝑏 squared. That expanded form is then 49 squared minus two times 49 times 𝑦 plus 𝑦 squared. From there, we can square 49 to get 2401. Negative two times 49 times 𝑦 equals negative 98𝑦. And then we can combine like terms, which will give us 2401 minus 98𝑦 plus two 𝑦 squared equals 1225.

At this point, we should recognize that we’re dealing with a quadratic. And that means to solve for 𝑦, we’re going to want to set this equation equal to zero. If we subtract 1225 from both sides, we get 1176 minus 98𝑦 plus two 𝑦 squared equals zero. Since all three of our coefficients are even, we can divide through by two, which gives us 588 minus 49𝑦 plus 𝑦 squared equals zero. It’s more common to write this in the form 𝑦 squared minus 49𝑦 plus 588. From here, you could solve with the quadratic formula or you could solve by factoring. We’re looking for two values that when multiplied together equal 588 and when added together equal negative 49. Those values will be negative 21 and negative 28. To solve finally, we’ll set each of these terms equal to zero. And then we find that 𝑦 equals 21 or 𝑦 equals 28.

Now that we have two possible values for 𝑦, we want to find our possible values for 𝑥. And to do that, we’ll take both values for 𝑦 and sub them into our third equation, which will give us 𝑥 equals 49 minus 21 or 𝑥 equals 49 minus 28. What we’re seeing is if 𝑦 is 21, 𝑥 equals 28. And if 𝑦 equals 28, 𝑥 equals 21. This means that one side must be equal to 28 centimeters and the other side must be equal to 21 centimeters. Additionally, we could check our answer by seeing if 21 plus 28 plus 35 is equal to 84, which it is.

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