In Exercises 47–48, solve each system by the method of your ch...

Question

In Exercises 47–48, solve each system by the method of your choice. (x + 2)/2 - (y + 4)/3 = 3 (x + y)/5 = (x - y)/2 - 5/2

Two variable linear equations for exercise 47 in college algebra, chapter 7.

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations. The first equation is x minus Y divided by seven equals X plus Y divided by eight minus three divided by two. And our second equation in our system is x minus three divided by two plus y minus three, divided by two equals negative five. And we get to solve this using any suitable method that we want. However, look at these equations. So before we can choose our suitable method, let's do a little bit of work on them and make them a little bit less fraction ng. And then we can figure out what method we want to use. So for equation one, doing some work there first to get rid of all the fractions, we're going to multiply all three of these fractions by the lowest common denominator, The lowest common denominator for 7, 8 and two is 56. So we are going to multiply all three of these by 56. And the seven in the 56 reduced to one and eight, the eight in the 56 reduced to one and seven And the two in the 56 reduced to one and 28. So now this will read eight times x minus y equals seven times X plus y -3 times 28 is 84. We can distribute the eight and the seven, so eight times x minus y is eight, X minus eight, Y equals seven times X plus Y is seven, X plus seven, Y minus 84. And then let's bring our X terms. And our y terms over to the left hand side of the equation eight X minus seven, X is just one. X negative eight. Y minus seven, Y is minus 15, Y equals the negative 84. And look how much nicer that looks. Let's do the same thing with equation number two. The common denominator here for 22 and one is two, multiplying all of these by two. R two's reduced to ones and then that too remains on the right hand side. So now it's just on the left, X minus three plus y minus three equals And on the right negative five times two is negative combining our like terms on the left, we've got X plus y negative three minus three is minus six equals negative 10. Let's bring that six over. We'll add six to both sides. So on the left we'll have x plus y equals on the right, will have negative 10 plus six is negative four. Well, look how much nicer that equation is. So now we can set up a snazzy daisy system of equations where the first one is x minus 15, Y equals negative 84. And the second one is x plus Y equals negative four. And I am going to use the addition method and I am going to multiply that whole first equation times a negative one and then add that to the second equation to get our X terms to cancel out. And we can solve for y. So negative one times X is negative X negative one times negative 15 wise plus 15 Y equals negative one times negative is positive 84. Now adding those together, X's cancel out one, Y plus 15 Y 16 Y equals negative four plus 84 is positive 80 Dividing both sides by 16. And we get y equals 5 divided by 16 is five. We've got our first answer, we've got our Y value. Now we can go back and plug this in for a Y. To solve for X. I like to go back as close to the original equation as possible in case I made any mistakes and just carried them through and for this I'm going to go back to the second equation where there's just one Y term and plug it in there. So rewriting this second equation will be x minus three divided by two plus instead of a Y. Up there. It will be a five minus three divided by two equals negative five. So just doing the work where we've got the blue where we've got new information five minus three is two. So now this is two over to bringing down the rest x minus three divided by two and 2/2 is one. So we've got x minus 3/2 plus one equals negative five. And we can subtract one from both sides and we'll have X -3/ Equals and -5 -1 is -6. Now we can multiply everything by two and we'll have the two's cancel out on the left hand side. So now I'm going to rewrite this up here, it's just x minus three equals negative six times two is negative 12. We can add three to both sides and we get X equals a negative nine and we have solved for X and for Y we look at our answer choices and negative 95 matches with answer choice C Well done, we'll catch on the next one.

In Exercises 47–48, solve each system by the method of your choice. (x + 2)/2 - (y + 4)/3 = 3 (x + y)/5 = (x - y)/2 - 5/2

Key Concepts

Systems of Linear Equations

Clearing Fractions in Equations

Methods for Solving Systems (Substitution, Elimination, or Graphing)

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