Question

Solve the following systems by Elimination

1) 4x+6y=32
3x-6y=3

2) -3+5y=-11
3x+7y=-1

Answer 1

Answer:

The solutions for both system of equations are as follows:

1. (5,2)
2. (2,-1)

Step-by-step explanation:

The first set of equations is:

$4x+6y=32\\3x-6y=3$

It can clearly be seen that the coefficients of y are already same in magnitude with different signs so we have to add both equations

So adding both equations, we get

$4x+6y+3x-6y = 32+3\\7x = 35\\\frac{7x}{7} = \frac{35}{7}\\x = 5$

Putting x=5 in equation 1

$4(5)+6y = 32\\20+6y = 32\\6y = 32-20\\6y = 12\\\frac{6y}{6} = \frac{12}{6}\\y = 2$

The solution is (5,2)

The second set of simultaneous equations is:

$-3x+5y=-113x+7y=-1$

We can see that the coefficients of x in both equations are same in magnitude with opposite signs so

Adding both equations

$-3x+5y+3x+7y = -11-1\\12y = -12\\\frac{12y}{12} = \frac{-12}{12}\\y = -1$

Putting y= -1 in first equation

$-3x+5(-1)=-11\\-3x-5=-11\\-3x=-11+5\\-3x=-6\\\frac{-3x}{-3} = \frac{-6}{-3}\\x = 2$

The solution is: (2,-1)

Hence,

The solutions for both system of equations are as follows:

1. (5,2)
2. (2,-1)