6x+4y=42 -3x+3y=-6 by elimination

To solve this system, we need to find x and y. First, let's find x. In the first equation (6x+4y=42), subtract 4y on both sides to isolate 6x. This will result in 6x=42-4y. Divide by 6 on both sides to isolate x and get x=7-[tex]\frac{2}{3}[/tex]y. Now we know what x equals. Go to the second equation ad plug this into it. So instead of -3x+3y=-6, you'll have -3(7-[tex]\frac{2}{3}[/tex]y)+3y=-6. Use the distributive property to get -21+2y+3y=-6. Add 2y and 3y to get -21+5y=-6. Add 21 to both sides to isolate 5y and get 5y=15. Divide by 5 on both sides to isolate y and get y=3. Now that we know that y=3, we can plug this back into the equation that was bolded above so that we can find x, so we'll have x=7-[tex]\frac{2}{3}[/tex](3) since y=3, and we'll get x=7-[tex]\frac{6}{3}[/tex], or x=7-2, or x=5.

x=5

y=2

drazo18 drazo18 · Answer 2 · 2020-12-03T07:28:35+01:00

Answer: (5, 3)

Step-by-step explanation:

Solve by Addition/Elimination

6x + 4y = 42 −3x + 3y = −6

Multiply each equation by the value that makes the coefficients of x opposite.

6x + 4y = 42

(2) ⋅ (−3x + 3y) = (2) (−6)

Simplify (2) ⋅ (−3x + 3y).

6x + 4y = 42

−6x + 6y = (2) (−6)

Multiply 2 by −6. 6x + 4y = 42

−6x + 6y = −12

Add the two equations together to e liminate x from the system.

6x + 4y = 42

±6x + 6y = −12

1 0y= 30

Divide each term in 10y = 30 by 10.

10y = 30

10 10

Cancel the common factor of 10.

y = 30/10

Divide 30 by 10.

y = 3

Substitute the value found for y into one of the original equations, then solve for x.

Substitute the value found for y into one of the original equations to solve for x. 6x + 4 (3) = 42

Multiply 4 by 3.

6x + 12 = 42

Move all terms not containing x to the right side of the equation.

6x = 30

Divide each term by 6 and simplify.

x = 5

The solution to the independent system of equations can be represented as a point.

(5, 3)

The result can be shown in multiple forms.

Point Form:

(5, 3)

Equation Form:

x = 5, y = 3