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

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A father’s age is 10 more than 2 times his son’s age. The sum of the squares of their ages is 4 more than 3 times the product of their ages. What are their ages?

05:10

### Video Transcript

A father’s age is 10 more than two times his son’s age. The sum of the squares of their ages is four more than three times the product of their ages. What are their ages?

We will begin by letting the ages of the father and son be 𝑦 years and 𝑥 years, respectively. We can then set up algebraic equations based on the information given in the question. Firstly, we are told that the father’s age is 10 more than two times his son’s age. Two times the son’s age is two 𝑥. 10 more than this is two 𝑥 plus 10. As this is the father’s age, we know that 𝑦 is equal to two 𝑥 plus 10. We will call this equation one.

Next, we are told that the sum of the squares of their ages is four more than three times the product of their ages. The product of their ages is 𝑥 multiplied by 𝑦, simply written 𝑥𝑦. Three times this is three 𝑥𝑦, and four more than this expression is three 𝑥𝑦 plus four. As this is equal to the sum of the squares of their ages, 𝑥 squared plus 𝑦 squared is equal to three 𝑥𝑦 plus four. We will call this equation two.

And we now have a pair of simultaneous equations. We can solve these by substitution. Our usual method here is to substitute an expression from the linear equation into the equation containing quadratic terms. We are going to substitute 𝑦 is equal to two 𝑥 plus 10 into equation two. This gives us 𝑥 squared plus two 𝑥 plus 10 all squared is equal to three 𝑥 multiplied by two 𝑥 plus 10 plus four. In order to square two 𝑥 plus 10, we write out two sets of parentheses. We multiply two 𝑥 plus 10 by two 𝑥 plus 10.

One way of distributing the parentheses is using the FOIL method, giving us four 𝑥 squared plus 20𝑥 plus 20𝑥 plus 100. Collecting like terms, this becomes four 𝑥 squared plus 40𝑥 plus 100. We can also distribute the parentheses on the right-hand side of our equation. This becomes six 𝑥 squared plus 30𝑥. And our equation is 𝑥 squared plus four 𝑥 squared plus 40𝑥 plus 100 is equal to six 𝑥 squared plus 30𝑥 plus four.

After simplifying the left-hand side, we can subtract five 𝑥 squared, 40𝑥, and 100 from both sides. This gives us a quadratic expression equal to zero. 𝑥 squared minus 10𝑥 minus 96 equals zero. We can solve this quadratic by factoring the right-hand side into two sets of parentheses with first term 𝑥. The second terms in our parentheses will have a product equal to the constant term negative 96 and a sum equal to the coefficient of 𝑥, in this case negative 10. One way of finding these values is to list the factor pairs of 96, and one such pair is 16 and six. This means that negative 16 multiplied by six is negative 96. Negative 16 plus six is equal to negative 10. Factoring 𝑥 squared minus 10𝑥 minus 96 gives us 𝑥 minus 16 multiplied by 𝑥 plus six.

When this expression equals zero, our equation has two possible solutions: 𝑥 equals 16 or 𝑥 equals negative six. Since we are dealing with ages in this question, our answers cannot be negative. This means that 𝑥 must be equal to 16. Substituting this value back into equation one, we have 𝑦 is equal to two multiplied by 16 plus 10. This is equal to 42. We can therefore conclude that the father’s age is 42 years and the son’s age is 16 years.

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