11) -x + y = -1 ; 2x - y = 0
y = -1 + x ; 2x - (-1+x) = 0 ⇒ 2x + 1 - x = 0 ⇒x = -1
y = -1 + (-1) ⇒ y = -2
12) -2x + y = -20 ; 2x + y = 48
y = -20 + 2x ; 2x + (-20 + 2x) = 48 ⇒ 2x -20 + 2x = 48 ⇒ 4x = 48 + 20
4x = 68 ⇒ x = 68/4 ⇒ x = 17
y = -20 + 2(17) ⇒ y = -20 + 34 ⇒ y = 14
13) 3x -y = -2 ; -2x + y = 3
y = 3 + 2x ; 3x - (3 + 2x) = -2 ⇒ 3x - 3 - 2x = -2 ⇒ x = -2 + 3 ⇒ x = 1
y = 3 + 2(1) ⇒ y = 3 + 2 ⇒ y = 5
14) x - y = 4 ; x - 2y = 10
x = 4 + y ; (4 + y) - 2y = 10 ⇒ 4 + y - 2y = 10 ⇒ 4 - y = 10
⇒ -y = 10 - 4 ⇒ -y = 6 ⇒ y = -6
x = 4 + (-6) ⇒ x = 4 - 6 ⇒ x = -2
15) x + 2y = 5 ; 3x + 2y = 17
x = 5 - 2y ; 3(5-2y) + 2y = 17 ⇒ 15 - 6y + 2y = 17 ⇒ -4y = 17 - 15
⇒ -4y = 2 ⇒ y = -2/4 ⇒ y = -1/2
x = 5 - 2(-1/2) ⇒ x = 5 + 2/2 ⇒ x = 5 + 1 ⇒ x = 6
Answer:
no
Step-by-step explanation:
The second purchase is exactly half-again larger than the first purchase, so sheds no light on the relative costs of the items. The per-item costs can be found when the ratio of items purchased is different from one buy to another.
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We can only say that the cost of 1 pear and 3 apples is $7.
Answer:
-13a^2 + 170ab - 13b^2.
Step-by-step explanation:
1. 36(a+b)^2 - 49(a-b)^2
This is the difference of 2 squares: a^2 - b^2 = (a + b)(a - b).
36(a+b)^2 - 49(a-b)^2
= (6(a + b) + 7(a - b))( 6(a + b) - 7(a - b))
= (6a + 7a + 6b - 7b)( 6a - 7a + 6b + 7b)
= (13a - b)(-a + 13b)
If you require the expansion of this it is:
-13a^2 + 170ab - 13b^2.
