Solve each problem using a system of equations in two variables. ...

Question

Solve each problem using a system of equations in two variables. See Example 6. Find two numbers whose sum is 17 and whose product is 42.

Accepted Answer

Hey, everyone. Here we are asked to find two numbers using a system of nonlinear equations in two variables. If they sum up to 64 yield a product of 295. When multiplied here, we have four answer choice options. Answer A five and 39 answer B five and 49 answer C five and 59 answer choice D five and 69. So here we need to begin by identifying our equations of our system of nonlinear equations. And so we need to assess our given problem statement. So first, we are told that we have two variables so we can choose the variables X and Y. And next, we are told that these two variables sum up to 64. This gives us our first equation where we now know that X plus Y is equal to 64. And now looking at the rest of our problem statement, we are also told that the two variables yield a product of 295 when multiplied. So here we get our second equation where we know that X times Y is equal to 295 So now we have identified our two equations in our system of equations. So now we need to begin solving. So we can first take our second equation where again we have X times Y is equal to 295. And now we want to solve for Y. So we simply divide both sides of the by X and we find that Y is equal to divided by X. So now with this expression that we found that describes our variable Y, we can utilize the substitution method by substituting in this expression for Y into our first equation. And so recalling our first equation, we have X plus Y is equal to 64. And now substituting we have X plus 295 divided by X is equal to 64. Now we just want to get X to be outside of this denominator in the fraction. So that means we simply need to multiply both sides of the equation by X. And so now multiplying by X, we have X squared plus is equal to 64 X. And now we just want the right side of our equation to be zero. So we simply subtract both sides to subtract X from both sides of our equation. So we now have X squared minus 64 X plus 295 is equal to zero. And now we want to start factoring and here we can factor by grouping. So we first want to split up our middle term of negative X into two different terms. So we have X squared minus 59 X minus five X plus 295 is equal to zero. Now, we just want to group together our first two terms and our last two terms. So we have the quantity of X squared minus 59 X minus the quantity of five X minus 295 and this is equal to zero. Now, we can notice that in our first set of parentheses, we can factor out X and in our second set of parentheses, we can factor out five. So factoring we have X times the quantity of X minus 59 minus five times the quantity of X minus 59 this is all still equal to zero. So now we see that both of our terms in our current equation have the common factor of this expression X minus 59. So now we want to factor out X minus 59 from both terms. So we have the quantity of X minus five times the quantity of X minus 59 this is equal to zero. Now, recalling the zero factor property, we can set each individual expression equal to zero and solve for our values of X. So starting with our first expression, we have X minus five. And again, because of the zero factor property, we can set this expression equal to zero. Now solving for X, we just add five to both sides and we find that X is equal to five. Now repeating this with our other expression, we have X minus 59 is equal to zero. Now solving for X, we just add 59 to both sides and we get X is also equal to 59. So now we have solved for our values of X. So we now need to use values to solve four values of Y. So recall our manipulated expression earlier, we have Y is equal to 295 divided by X. And now we can utilize this equation by substituting in the values we found for X. And so we can solve directly for values of Y. So starting with our first value of X where we have X is equal to five, we substitute in this value. So we have Y is equal to divided by five. Now simplifying we find that Y is equal to 59. So we see that when X is equal to five, that Y is equal to 59 repeating this with the other value where X is equal to 59 we now have Y is equal to 295 divided by 59. And now simplifying, we get Y is equal to five. So we see also here that when X is equal to 59 that Y is equal to five. Now here looking at the values we obtained for X and Y, we can see that X and Y are the same numbers when they are interchanged. So when X is equal to five, Y is equal to 59 or when Y is equal to five, X is equal to 59. So this tells us that the numbers that we were looking for are five and 59. So with these two values, we are left with answer choice. C where again our values are five and 59. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each problem using a system of equations in two variables. See Example 6. Find two numbers whose sum is 17 and whose product is 42.

Key Concepts

System of Equations

Linear Equations

Factoring Quadratics

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