Answer:
(-2, -3)
(3, 12)
Step-by-step explanation:
To solve this, we're gonna get rid of the y's with substitution
x² + 2x - 3 = 3x + 3
Let's make this equation equal to zero
Subtract 3 from both sides
x² + 2x - 3 = 3x + 3
- 3 - 3
x² + 2x - 6 = 3x
Subtract 3x from both sides
x² + 2x - 6 = 3x
- 3x - 3x
x² - x - 6 = 0
Factor the equation
(x - 3)(x + 2) = 0
This means x can be -2 or 3
Let's solve it with -2 first, plug the new x in y = 3x + 3
y = 3(-2) + 3
y = -6 + 3 = -3
Do the same for x = 3
y = 3(3) + 3
y = 9 + 3 = 12
Step-by-step explanation:
sol;
x+1=y...(1)
3y-7=2x....(2)
or, 3(x+1)-72x [from (1)]
or, 3x+3-7=2x
or, 3x-2x=7-3
x=4
now,
putting the value of x in (1)
y=x+1
=4+1
=5
PR and SQ are the diagonal of PQRS.
Without LCD :
1/2(4x + 6) = 1/3(9x - 24)
2x + 3 = 3x - 8
2x - 3x = -8 - 3
-x = - 11
x = 11
with LCD :
1/2(4x + 6) = 1/3(9x - 24) ...multiply by 6
3(4x + 6) = 2(9x - 24)
12x + 18 = 18x - 48
12x - 18x = -48 - 18
-6x = - 66
x = -66/-6
x = 11
Yes, the answers are the same :)