Y = 5x + 5
y - x = 5
You can use substitution to solve this system. Use y's expression (5x + 5) for y in the second equation to solve for x:
y - x = 5
5x + 5 + x = 5
6x + 5 = 5
6x = 0
x = 0
Substitute your value for x into one of the original equations to y:
y = 5x + 5
y = 5(0) + 5
y = 5
Finally, substitute both values into both original equations to check your work:
5 = 5(0) + 5 --> 5 = 5 <--True
5 - 0 = 5 --> 5 = 5 <--True
Answer:
x = 0
y = 5
You would write 5/3 for it to be an improper fraction
C
substitute a = - 1 into the expression
- (- 1) - 12(- 1)² = 1 - (12 × 1 ) = 1 - 12 = - 11 → C
So with this, I will be using the substitution method. With the first equation, substitute (y+3) into the x variable and solve for y:
Next, now that we have the value of y, substitute it into either equation to solve for x:
<u>And this is how you get your final answer (5,2).</u>
