Question Video: Adding Fractions with Integer or Linear Agebraic Expressions for Numerators and Different Linear Algebraic Expressions for Denominators | Nagwa Question Video: Adding Fractions with Integer or Linear Agebraic Expressions for Numerators and Different Linear Algebraic Expressions for Denominators | Nagwa

Question Video: Adding Fractions with Integer or Linear Agebraic Expressions for Numerators and Different Linear Algebraic Expressions for Denominators Mathematics

Write 4/(π‘₯ + 1) + 2π‘₯/(2π‘₯ + 3) as a single fraction in its simplest form.

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Video Transcript

Write four over π‘₯ plus one plus two π‘₯ over two π‘₯ plus three as a single fraction in its simplest form.

To solve this problem, we can see that we’ve actually got two algebraic fractions and we want to add them. And as with any type of fraction, when you want to add fractions, you need to find a common denominator. And one method we can actually use is the cross multiplication method, which allows us to add our fractions.

So what we have is if we have two fractions π‘Ž over 𝑏 plus 𝑐 over 𝑑, then we multiply π‘Ž by 𝑑 β€” so we get π‘Žπ‘‘ β€” plus 𝑐 multiplied by 𝑏 β€” so 𝑐𝑏. And then you multiply the two denominators β€” so 𝑏 by 𝑑. So we get 𝑏𝑑. Okay, fab, so this is actually a method. It’s gonna help us to solve our problem. So let’s get on add together our algebraic fractions.

So we can say that four over π‘₯ plus one plus two π‘₯ over two π‘₯ plus three is gonna be equal to four multiplied by two π‘₯ plus three plus two π‘₯ multiplied by π‘₯ plus one cause again we cross multiplied all over π‘₯ plus one multiplied by two π‘₯ plus three. Okay, great, so we’ve now completed the first step.

So now, what we’re gonna do is actually expand the parentheses on the numerator. So we’re gonna get eight π‘₯ as we got four multiplied by two π‘₯ plus 12 because we have four multiplied by three plus two π‘₯ squared cause two π‘₯ multiplied by π‘₯ gives us two π‘₯ squared plus two π‘₯ and then this is all over π‘₯ plus one multiplied by two π‘₯ plus three.

So therefore, what we can do now is actually simplify our numerator. So we get two π‘₯ squared plus 10π‘₯ plus 12 over π‘₯ plus one multiplied by two π‘₯ plus three. And we got that because we actually combine the like terms that we could. So we had eight π‘₯ plus two π‘₯ give us 10π‘₯. Okay, great, so we’ve now written it as a single fraction. But is it in its simplest form?

Well, actually, I think there’s one more step we can do because we can actually take a factor out of our numerator. And this gives us two multiplied by π‘₯ squared plus five π‘₯ plus six because two is a factor of each of our terms over π‘₯ plus one multiplied by two π‘₯ plus three. But is this the final answer? Well, actually, if we look at this now, we can actually see that we’ve got in the numerator a quadratic that we can actually factor. Well, let’s have a quick look at how we’re gonna do that.

So we’ve got π‘₯ squared plus five π‘₯ plus six. Well, we’ve got positive π‘₯ squared, which means that we’re gonna have an π‘₯ at the beginning of each of our parentheses. Okay, so now what? Well, what we need to do is we need to find a pair of factors that multiply together to give us positive six and add together to give us positive five. So therefore, the two numbers that we’re gonna get are positive three and positive two because two multiplied by three gives us six and two add three gives us five or positive five.

So great, we found our factors. They’re π‘₯ plus three multiplied by π‘₯ plus two. So therefore, we can say that if we write four over π‘₯ plus one plus two π‘₯ over two π‘₯ plus three as a single fraction in its simplest form, it’s gonna be equal to two multiplied by π‘₯ plus two π‘₯ plus three over π‘₯ plus one two π‘₯ plus three.

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