How do I solve this system of equations?

Question

3x-2y=4
4x-5y=-4

How would I go about solving the above system of equations problem? I think I know what the answer is, but I want to make sure I am doing this right.

A easy-to-follow explanation of the steps you take would be really helpful. Thanks in advance, I really appreciate it!

Answer 1

It's not hard. Solve for x (or y, it's your choice) in the first equation, which gives you x=(4-2y)/3. Then plug this into x in the second equation, 4((4-2y)/3)-5y=-4. Then solve for y, and plug this value into y again to find x. Vice versa if you solved for y.

Another way to solve this:

Another way to rewrite the second equation is -(4x-5y) = 4, right? Since both equations now = 4, you can set the equations equal to each other, and solve from there. 3x-2y = -(4x-5y)

Answer 2

There are a lot of ways to solve systems, but my students seem to like elimination best. Multiply both rows by some number that will allow you to eliminate one of the variables when you add the two rows.

In the system...
3x - 2y = 4
4x - 5y = -4
...you could either multiply the top by 4 and the bottom by -3 to get rid of the x's, or you could multiply the top by 5 and the bottom by -2 to get rid of the y's.

4(3x - 2y) = 4(4)
-3(4x - 5y) = -3(-4)

12x - 8y = 16
-12x + 15y = 12

Adding the rows yields
7y = 28
y = 4

Then substitute into any equation and solve for x.
3x - 2y = 4
3x - 2(4) = 4
3x - 8 = 4
3x = 12
x = 4

Your solution is x = 4 and y = 4.

Answer 3

If you solve the first equation for x, you get

3x = 4 + 2y
x = 4/3 + 2y/3

Plug this into the second equation, and you get

4(4/3 + 2y/3) -5y = -4
16/3 + 8y/3 - 15y/3 = -12/3
-7y/3 = -28/3
y=4

plug this value into the equation that you got for x:
x = 4/3 + 2y/3
x = 4/3 + 2(4)/3
x = 4/3 + 8/3
x = 4

Now, check you answer by plugging in both values for x and y into the original 2 equations:

3x-2y=4
3(4) - 2(4) = 4
12 - 8 = 4

4x-5y=-4
4(4) - 5(4) = -4
16 - 20 = -4

Answer 4

x=y=4

Answer 5

There are several methods Here is one

eqn A is 3x-2y=4
eqn B is 4x-5y=4

We make coefficient of x equal
eqn Ax4 => 4[3x-2y=4]
=> 12x-8y=16
eqn Bx3 => 3[4x-5y=4]
=> 12x-15y=12
subtract the two equations that have x coeff equal
12x-8y =16
12x-15y= 12
- + -
-------------------------------
7y=28
y=28/7 =4

put this value in eqn A (or u could use eqn B) and get the value of x


3x -2*4=4
3x-8=4
3x=4+8
x=12/3 =4

ans x=4, y=4

Answer 6

You want to be able to compare the equations to each other so you can eliminate one of the variables. Let's try to eliminate the y's....

Multiply the first equation by 5:
15x - 10y = 20
Multiply the second equation by 2:
8x - 10y = -8
Now subtract the second equation from the first:
7x - 0y = 28
So x = 4. Substitute that into either equation (we'll use the first):
3(4) - 2y = 4
12 - 2y = 4
y = 4

So x=4, y=4.

Answer 7

let 3x-2y=4 be x=(4+2y)/3
and let x=(-4+5y)/4

equate both:
(4+2y)/3=(-4+5y)/4

multiply both sides by the denominator of the other side(and cancel out the denominators)
16+8y=-12+15y
goes to 7y=28
therefore y=4

and x=(4+2(4))/3=4

Answer 8

3x - 2y = 4
4x - 5y = -4

Multiply top by 5 and bottom by -2

15x - 10y = 20
-8x + 10y = 8

7x = 28
x = 4

3x - 2y = 4
3(4) - 2y = 4
12 - 2y = 4
-2y = -8
y = 4

ANS : x = 4 and y = 4

Answer 9

5x + 2y = 15 -5x - 2y = -5 _____________ 0 = 10 you remedy employing combining words, you upload the two equations. 0=10 ability this technique of equation has no answer in any respect, via fact the slopes of the two strains are a similar, in different words are parallel strains, so there is no longer factor of interception

Answer 10

U go about solving the eqns by eliminating one of the variables. For this, u first make the co-efficients of one variable(say x) equal in both eqns. So multiply the 1st eqn by 4 & the 2nd eqn by 3.
This gives us

12x-8y=16................(iii)
12x-15y= -12............(iv)

Now, to eliminate x, subtract eqn (iv) from eqn (iii), i.e perform the operation (iii)-(iv).

This gives the result

7y=28.....................(v)

Dividing both sides of (v) by 7, we get

y=4.

Substituting this value of y in the first eqn, we get

x=4.

U can verify the solution we obtained by substituting x=4, y=4 in the second eqn, which gives us the net result of -4 on the Left Hand Side(LHS), which is equal to the Right Hand Side(RHS)...... Done!

Hope this helps... bye!