How do you solve these linear systems using any algebraic method: 3x-4y=16 / 5x+6y=14?

Question

THANK YOU SO MUCH.

Answer 1

Let us try to eliminate one variable.
So, multiply the 1st equation by 3 and 2nd equation by 2 and
you will get,
9x - 12y = 48
10x + 12y = 28

Now add the equations to eliminate y.

9x + 10x - 12y + 12y = 48 + 28
19x = 76
x = 4

y = 1/6 (14 - 5x)
y = 1/6 (-6)
y = -1

(4, -1) is the solution.

Answer 2

Algebraic method is applied to solve system of linear equations

Step1 => Bring the equations into y=mx+c form (slope intercept form)
3x-4y=16 becomes y = (3/4)x - 4
5x+6y=14 becomes y = (-5/6)x+7/3

Step2 => Now equate both the equations
(3/4)x - 4 = (-5/6)x+7/3
=> (3/4 + 5/6)x = 4 + 7/3
=> (19/12)x = 19/3
=> x = 4

Step3 => Now substitue this value in any of the slope intercept forms to get y
y = (3/4)x - 4
=> y = (3/4)4 - 4
=> y = -1

Thus the solution is (4,-1) for the given linear systems.

Answer 3

3x - 4y = sixteen (eq1) 5x - 6y = 14 (eq2) 2 techniques, substitution and addition (removing). Substitution: remedy between the equations for between the variables. i will remedy eq1 for y. 3x - 4y = sixteen 4y = 3x - sixteen y = 3x/4 - 4 substitute y = 3x/4 - 4 into the eq2: 5x - 6y = 14 5x - 6(3x/4 - 4) = 14 5x - 18x/4 + 24 = 14 remedy for x: 2x/4 = -10 x/2 = -10 x = -20 Plug this value into y = 3x/4 - 4 y = 3(-20)/4 - 4 y = 3(-5) - 4 y = -15 - 4 y = -19 removing: 3x - 4y = sixteen (eq1) 5x - 6y = 14 (eq2) Multiply eq1 by utilising 5 and eq2 by utilising -3: 15x - 20y = 80 -15x + 18y = -40 two upload the equations mutually: -2y = 38 y = -19 substitute this value for y into eq1 or eq2: 3x - 4y = sixteen 3x - 4(-19) = sixteen 3x + seventy six = sixteen 3x = -60 x = -20

Answer 4

9x - 12y = 48
10x + 12y = 28---ADD
19x = 76
x = 4

20 + 6y = 14
6y = - 6
y = - 1

x = 4 , y = - 1