Multiply each equation by a number that produces opposite coefficients for x or y. 4x + 5y = 7 3x-2y=-12

Step-by-step explanation:

We could multiply the first equation by -3 and, separately, multiply the second equation by 4.  The result would be:

-12x - 15y = -21

12x  - 8y = -48

the x terms now cancel.  The result is:

- 15y = -21

- 8y = -48

-23y = -69, or y = 3.  If y = 3, then 3x - 2y = -12 becomes:

3x - 2(3) = -12, or

3x - 6 = -12, or    3x = -6, and so x = -2.

The solution is (-2, 3).

-1

Step-by-step explanation:

4x+5y=7

3x-2y= -12

To get the opposite coefficient, we multiply by -1

-1(4x+5y=7 )

-4x-5y = -7

-1*(3x-2y= -12)

-3x +2y = 12

Both equations now have the opposite coefficients on both x and y.

1st option is: -3

2nd option is: 8

Step-by-step explanation: I just did the assessment on edge

Let's solve your system by substitution.

3x−2y=−12;4x+5y=7

Step: Solve3x−2y=−12for x:

3x−2y=−12

3x−2y+2y=−12+2y(Add 2y to both sides)

3x=   2y−12

3          3

=

2y−12

3

(Divide both sides by 3)

x=

2

3    y−4

Step: Substitute

2

3

y−4 for x in 4x+5y=7:

4x+5y=7

4( 2/3  y−4)+5y=7

23

3

y−16=7(Simplify both sides of the equation)

23

3

y−16+16=7+16(Add 16 to both sides)

23

3

y=23

23

3  y

23

3      =  23

23

3

(Divide both sides by 23/3)

y=3

Step: Substitute 3 for y in x=

 2  

3 y−4:

x=

 2  

3 y−4

x=    2  

  3 (3)−4

x=−2(Simplify both sides of the equation)

x=−2 and y=3                    (underlined and / are fractions or division)

Step-by-step explanation:

The coefficient of x needs to be 2.

The current coefficient is 2/5.

To eliminate the 5 in the denominator, we can multiply the equation by 5.

We get

Now, to solve the equations, all we need to do is add them.

Hope this helps!

The solution is

Solution:

Given system of equations are:

4x + 6y = 7 --------- eqn 1

3x - 2y = -12 -------- eqn 2

Let us solve the system of equations by elimination method

Multiply eqn 2 by 3

9x - 6y = -36 ----- eqn 3

Add eqn 1 and eqn 3

4x + 6y = 7

9x - 6y = -36

( + ) --------------

13x = -29

Substitute in eqn 1

Thus the solution is

x = -2, y = 3

Step-by-step explanation:

Given equations are,

4x + 5y = 7 (1)

3x - 2y = -12 (2)

Multiply equation (1) by 3

12x + 15y = 21 (3)

Multiply equation (2) by (-4)

-12x + 8y = 48 (4)

Now add equations (3) and (4),

(12x + 15y) + (-12x + 8y) = 21 + 48

(12x - 12x) + (15y + 8y) = 69

23y = 69

y = 3

From equation (2),

3x - 2(3) = -12

3x - 6 = -12

3x = 6 - 12

x = -2

Therefore, x = -2 and y = 3 will be the answer.

The solution is (x, y) = (-2, 3)

Solution:

Given system of equations are:

4x + 5y = 7 eqn 1

3x -2y = -12 eqn 2

Multiply eqn 1 by 2

8x + 10y = 14 eqn 3

Multiply eqn 2 by 5

15x - 10y = -60 eqn 4

Add eqn 3 and eqn 4

8x + 10y = 14

15x - 10y = -60

( + )

23x = -46

x = -2

Substitute x = -2 in eqn 1

4(-2) + 5y = 7

-8 + 5y = 7

5y = 15

y = 3

Thus the solution is (x, y) = (-2, 3)

To produce opposite x terms:
-3(4x + 5y = 7) ⇒ -12x -15y = -21
4(3x - 2y = -12) ⇒ 12x - 8y = -48
so if you add the equations the x terms will cancel out

To produce opposite y terms:
2(4x + 5y = 7) ⇒ 8x + 10y = 14
5(3x - 2y = -12) ⇒ 15x -10y = -60
so if you add the equations now the y terms will cancel out

What you are doing is finding the LCM (least common multiple) of the coefficients for the terms you want to cancel out...
The LCM for 4 and 3 (coefficients of x) is 12 so you multiple each by what is required to get -12 and +12.
The LCM for 5 and 2 (coefficients of y) is 10 so you multiple each by what is required to get -10 and +10.