Hi Angelica;
3x+7y=16 and 9x+5y=16
The coefficients of x are 3 and 9. These should be identical so we may apply the practice of elimination.
Let's take the first equation...
3x+7y=16
Let's multiply both sides by 3...
3(3x+7y)=(16)(3)
9x+21y=48
Let's subtract this from the second equation...
9x+5y=16
(9x+21y=48)
016y=64
The x is eliminated.
16y=64
(16y)/16=64/16
y=4
Let's plug this into either equation to establish the value of x. I randomly select the first...
3x+7y=16
3x+[(7)(4)]=16
3x28=16
Let's add 28 to both sides...
3x28+28=16+28
3x=12
Let's divide both sides by 3...
(3x)/3=12/3
x=4
Let's plug both values into the second equation to verify results...
9x+5y=16
[(9)(4)]+[(5)(4)]=16
A negative number multiplied by a negative number has a positive result...
3620=16
16=16
Feb 3

Vivian L.