In Exercises 19–30, solve each system by the addition method. 3x = 4y + 1 3y = 1 - 4x

Question

Exercise 29: Solve the system of equations 3x = 4y + 1 and 3y = 1 - 4x using the addition method.

Accepted Answer

Hey everyone welcome back in this problem. We are asked to consider the given system of equations and solve using the addition method. And the equations were given as five X equals 19 minus six Y. And six X plus 26 equals five Y. Okay so we have a system of two equations. Let's go ahead and just label these equations were gonna call equation + X equals 19 minus six Y. An equation two is going to be six X plus 26 equals five Y. Okay, we're told to use the addition method. We're using the addition method. We need the equations to be in the same form. Okay so equation one you can see the X term is on the left hand side. The constant term in the white term on the right hand side and an equation to we have the X. Term and the constant term on the left, the Y term on the right. So let's go ahead and write equation too. So that is in the same form as equation one. Okay, so we'll get to Star. Okay so we're just calling it to Star because it's the exact same equation, it's just been rearranged and it's gonna be six X. We're gonna move the 26 to the right hand side is equal to minus 26 plus five Y. Okay, so now it looks the same as equation one in terms of the form and we can move on with our edition method. Now when we're doing the addition method we want to add the two equations. Okay to eliminate one of the variables So we can choose to eliminate the variable X. Or we can choose to eliminate the variable Why? Um in this case let's go ahead and choose X. Now in order to add the equations and eliminate one of the variables, we need the coefficient in front of them to be the same but with an opposite sign. Okay well we have five X. And we have six X. So in order to make those the same, we're gonna have to maybe turn them into 30 X. Okay, that would be the easiest. We multiply the first equation by six. We multiply the second equation by minus five. Okay, so let's do that. Let's take equation one And multiply it by six. And then we're going to add that's our addition method Equation two. And we're gonna use equation to star because it's in the form we want we're gonna multiply it by -5. Alright, so starting with equation one time 6, 30 x is equal to -36. Why? And then equation to start? Times minus five. Okay, so six X. Times minus five is going to give us minus 30 X. And this is what we wanted. We wanted the same coefficient with different signs. Okay, on the right hand side we get -25. Why? Alright, now we add them together. That's our addition method and we're going to see that the 30 X plus negative 30 X. Gives us zero. So we've eliminated the X term. Sometimes the addition method is called the elimination method for this reason. Okay 114 plus 130 is gonna give us 244 and then minus 36 plus negative 25 Y. Well that's going to give us minus 61. Y. Now we only have wise in constant terms, so we can go ahead and solve for y moving the 61 Y to the left hand side. We get 61 Y is equal to 244 and dividing by 61 we get Y is equal to four. Okay? So we know that Y equals four is the y value that solves the system. Now we need to find the corresponding X value. Okay we need an X or an X value and a Y value in order to solve the system. So using our addition method we found y equals four. Now we can go ahead and because we're solving both equations, both equations have to be satisfied. We can sub y equals four into either equation Okay in this case we're going to choose to sell it into equation one. So sub y equals four into equation one. But you could choose equation to if you wanted and we are going to get five X is equal to 19 minus six times Y. And we found that Y is equal to four. so minus six times four. We're gonna have 5x. is equal to 19-24. Five X is equal to minus five and dividing by five X is equal to minus one. Okay, So now we have an X. Value and a Y. Value and so we can write our solution set as the point that we found minus one four. So the solution to the system is -14. That's going to correspond with answer b. Thanks everyone for watching. See you in the next video.

In Exercises 19–30, solve each system by the addition method. 3x = 4y + 1 3y = 1 - 4x

Verified Solution