A school is holding a carnival and hopes to raise $500. Child tickets cost $3 and adult tickets cost $5. If the school sells x c
2 answers:
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a. (3,4) is only the solution to Equation 1.
(4, 2.5) is the solution to both equations
(5,5) is the solution to Equation 2
(3,2) is not the solution to any equation.
b. No, it is not possible to have more than one (x,y) pair that is solution to both equations
Step-by-step explanation:
a. Decide whether each neither of the equations,
i (3,4)
ii. (4,2.5)
ill. (5,5)
iv. (3,2)
To decide whether each point is solution to equations or not we will put the point in the equations
Equations are:
Equation 1: 6x + 4y = 34
Equation 2: 5x – 2y = 15
<u>i (3,4) </u>
Putting in Equation 1:
Putting in Equation 2:
<u>ii. (4,2.5)</u>
Putting in Equation 1:
Putting in Equation 2:
<u>ill. (5,5)</u>
Putting in Equation 2:
<u>iv. (3,2)</u>
Putting in Equation 2:
Hence,
(3,4) is only the solution to Equation 1.
(4, 2.5) is the solution to both equations
(5,5) is the solution to Equation 2
(3,2) is not the solution to any equation.
b. Is it possible to have more than one (x, y) pair that is a solution to both
equations?
The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution
So,
a. (3,4) is only the solution to Equation 1.
(4, 2.5) is the solution to both equations
(5,5) is the solution to Equation 2
(3,2) is not the solution to any equation.
b. No, it is not possible to have more than one (x,y) pair that is solution to both equations
Keywords: Linear equations, Ordered pairs
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