1 the knitting club sold 40 scarves and hats at a winter festival and made $700 from the sales. they charged…

1 the knitting club sold 40 scarves and hats at a winter festival and made $700 from the sales. they charged $18 for each scarf and $14 for each hat. if s represents the number of scarves sold and h represents the number of hats sold, which system of equations represents the constraints in this situation? a {40s + h = 700, 18s + 14h = 700} b {18s + 14h = 40, s + h = 700} c {s + h = 40, 18s + 14h = 700} d {40(s + h) = 700, 18s = 14h} 2 here are two equations: equation 1: 6x + 4y = 34 equation 2: 5x - 2y = 15 a. decide whether each (x, y) pair is a solution to one equation, both equations, or neither of the equations. i. (3, 4) ii. (4.2, 2.5) iii. (5, 5) iv. (3, 2) b. is it possible to have more than one (x, y) pair that is a solution to both equations? explain or show your reasoning.

1 the knitting club sold 40 scarves and hats at a winter festival and made $700 from the sales. they charged $18 for each scarf and $14 for each hat. if s represents the number of scarves sold and h represents the number of hats sold, which system of equations represents the constraints in this situation? a {40s + h = 700, 18s + 14h = 700} b {18s + 14h = 40, s + h = 700} c {s + h = 40, 18s + 14h = 700} d {40(s + h) = 700, 18s = 14h} 2 here are two equations: equation 1: 6x + 4y = 34 equation 2: 5x - 2y = 15 a. decide whether each (x, y) pair is a solution to one equation, both equations, or neither of the equations. i. (3, 4) ii. (4.2, 2.5) iii. (5, 5) iv. (3, 2) b. is it possible to have more than one (x, y) pair that is a solution to both equations? explain or show your reasoning.

Answer

Explanation:

Step1: Analyze problem 1

The total number of items sold is 40, so $s + h=40$. The total money made is $700$, with scarves at $18$ each and hats at $14$ each, so $18s + 14h=700$.

Step2: Analyze problem 2 - i

For $(3,4)$ in Equation 1: $6\times3 + 4\times4=18 + 16=34$. In Equation 2: $5\times3-2\times4 = 15 - 8=7\neq15$. So it's a solution to Equation 1 only.

Step3: Analyze problem 2 - ii

For $(4.2,2.5)$ in Equation 1: $6\times4.2+4\times2.5 = 25.2+10=35.2\neq34$. In Equation 2: $5\times4.2-2\times2.5=21 - 5 = 16\neq15$. So it's a solution to neither equation.

Step4: Analyze problem 2 - iii

For $(5,5)$ in Equation 1: $6\times5+4\times5=30 + 20=50\neq34$. In Equation 2: $5\times5-2\times5=25 - 10 = 15$. So it's a solution to Equation 2 only.

Step5: Analyze problem 2 - iv

For $(3,2)$ in Equation 1: $6\times3+4\times2=18 + 8=26\neq34$. In Equation 2: $5\times3-2\times2=15 - 4 = 11\neq15$. So it's a solution to neither equation.

Step6: Analyze problem 2 - b

Two non - parallel lines (represented by linear equations $6x + 4y=34$ and $5x-2y=15$) intersect at exactly one point. So there is only one $(x,y)$ pair that is a solution to both equations.

Answer:

  1. C

i. A solution to Equation 1 only ii. A solution to neither equation iii. A solution to Equation 2 only iv. A solution to neither equation b. No. Two non - parallel lines represented by linear equations intersect at exactly one point.