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# solve 2x + 3y = 7, x+y=3

+3
6204
3

solve 2x + 3y = 7, x+y=3

Aug 28, 2014

#1
+3450
+8

There are three ways to solve questions like these.

We can solve this by substitution, which is where you find what, say, x equals, and put that in for x in the other equation to solve for y. Then put that number in for y and solve for x.

We can solve by elimination, which is where you set it up so when you "add" these two equations together, you will get one varible, (so just x, or just y) and you can solve for that, then put that in and solve for the other varible.

Lastly you can solve by graphing. You can solve these by setting these lines up in the slope intercept form, graph the two lines, and then where they intersect is solution to this system.

Let's solve this by substitution. You can solve it any way you'd like, but this is how I'll solve this.

Let's find what x equals in the second equation, which is

x+y=3        ---Subtract y from both sides

x = 3 - y

Now we can put what x equals in the first equation, and find the value of y.

2x + 3y = 7              ---Put (3 - y) in for x, because this is what x equals

2(3-y) + 3y = 7           ---Distribute out 2(3-y)

6-2y+3y=7            ---Combine -2y and 3y to get y

6+y=7                  ---Subtract 6 from both sides

y = 1

Now we know the value of y. Let's put this in the first equation and solve for x.

2x + 3y = 7

2x + 3(1) = 7

2x + 3 = 7           ---Subtract  from both sides

2x = 4              ---Divide both sides by 2

x = 2

The solution to this system of equations is, *drumroll please*, x = 2 and y = 1!

Aug 28, 2014

#1
+3450
+8

There are three ways to solve questions like these.

We can solve this by substitution, which is where you find what, say, x equals, and put that in for x in the other equation to solve for y. Then put that number in for y and solve for x.

We can solve by elimination, which is where you set it up so when you "add" these two equations together, you will get one varible, (so just x, or just y) and you can solve for that, then put that in and solve for the other varible.

Lastly you can solve by graphing. You can solve these by setting these lines up in the slope intercept form, graph the two lines, and then where they intersect is solution to this system.

Let's solve this by substitution. You can solve it any way you'd like, but this is how I'll solve this.

Let's find what x equals in the second equation, which is

x+y=3        ---Subtract y from both sides

x = 3 - y

Now we can put what x equals in the first equation, and find the value of y.

2x + 3y = 7              ---Put (3 - y) in for x, because this is what x equals

2(3-y) + 3y = 7           ---Distribute out 2(3-y)

6-2y+3y=7            ---Combine -2y and 3y to get y

6+y=7                  ---Subtract 6 from both sides

y = 1

Now we know the value of y. Let's put this in the first equation and solve for x.

2x + 3y = 7

2x + 3(1) = 7

2x + 3 = 7           ---Subtract  from both sides

2x = 4              ---Divide both sides by 2

x = 2

The solution to this system of equations is, *drumroll please*, x = 2 and y = 1!

#2
+95361
+3

Aug 29, 2014
#3
+8263
0

My mouth is opened! NinjaDevo is really smart, unlike me.

Btw, I like the way he answered it!

Aug 29, 2014