Question

ONE OF THE VERY LAST QUESTION! Solve the system of equations.
4x + 7y = 6 and 6x + 5y = 20

(5, 2)
(-5, 2)
(5, -2)
no solution
infinitely many solutions

Answer 1

There are two basic ways to solve this type of problem. I will go over both ways using an example, so that you can follow the steps with your problem.

I will use\[5x+6y=4\]\[4x+9y=14\]The first way is fairly simple; just solve one of the equations for one variable. I will solve the first equation for x.\[5x+6y=4\]\[5x=4-6y\]\[x=\frac{ 4 }{ 5 }-\frac{ 6y }{ 5 }\]Now that I've solved for x in terms of y for the first equation, I can substitute that in for x in the second equation.\[4x+9y=14\]\[4(\frac{ 4 }{ 5 }-\frac{ 6y }{ 5 })+9y=14\]\[\frac{ 16 }{ 5 }-\frac{ 24y }{ 5 }+\frac{ 45y }{ 5 }=\frac{ 70 }{ 5 }\](Note that I have multiplied 9y and 14 by 5 each to make it slightly easier to solve).\[\frac{ 21y }{ 5 }=\frac{ 70 }{ 5 }-\frac{ 16 }{ 5 }=\frac{ 54 }{ 5 }\]\[21y=54\]\[y=\frac{ 54 }{ 21 }=\frac{ 18 }{ 7 }\]Now that I know what y is, I can go back to my first equation. Here, I just plug in this value of y I found, and this will give me x.\[x=\frac{ 4 }{ 5 }-\frac{ 6 }{ 5 } \times y\]\[x=\frac{ 4 }{ 5 }-\frac{ 6 }{ 5 } \times \frac{ 18 }{ 7 }\]\[x=\frac{ 4 }{ 5 }-\frac{ 6 \times 18 }{ 5 \times 7 }=\frac{ 4 \times 7 }{ 5 \times 7 }-\frac{ 6 \times 18 }{ 5 \times 7 }=\frac{ 4 \times 7 - 6 \times 18 }{ 5 \times 7 }=\frac{ -80 }{ 35 }=\frac{ -16 }{ 7 }\]So y=18/7 and x=-16/7

The other way is by elimination. Our equations were\[5x+6y=4\]\[4x+9y=14\]The thing we want to do here is to add or subtract the two equations from one another in order to eliminate one of the variables. Let's eliminate y. In order to do this, we're going to multiply the second equation by 6/9 and then subtract it from the first equation. Like this:\[5x+6y-\frac{ 6 }{ 9 }(4x+9y)=4-\frac{ 6 }{ 9 } \times 14\]\[5x+6y-\frac{ 8 }{ 3 }x-6y=4-\frac{ 28 }{ 3 }\]\[\frac{ 15x }{ 3 }-\frac{ 8x }{ 3 }=\frac{ 12 }{ 3 }-\frac{ 28 }{ 3 }\]\[\frac{ 7x }{ 3 }=\frac{ -16 }{ 3 }\]\[x=\frac{ -16 }{ 7 }\]Now that we know what x is, we just plug it into either equation to find y. I'm going to plug it into the second equation.\[4x+9y=14\]\[4 \times \frac{ -16 }{ 7 }+9y=14\]\[\frac{ -64 }{ 7 }+\frac{ 63y }{ 7 }=\frac{ 98 }{ 7 }\]\[\frac{ 63y }{ 7 }=\frac{ 98+64 }{ 7 }=\frac{ 162 }{ 7 }\]\[63y=162\]\[y=\frac{ 162 }{ 63 }=\frac{ 18 }{ 7 }\]And so as you can see, you get the same answer either way.

Now, you do also mention infinitely many solutions. This can ONLY happen if you don't have as many unique equations as you do variables. So if you had two equations, but one was a multiple of the other, you would get infinitely many solutions.

You would get no solutions if you had an impossible system. An example would be\[4x+3y=7\]\[4x+3y=1\]I hope that helps!