Step-by-step explanation:
x + y = 16 (*)
x^2 + y^2 = 146 (**)
(*) <=> x= 16-y
substitute x = 16 - y
(**) <=> (16-y)^2 + y^2 = 146
<=> y^2 - 2.16.y + 16^2 + y^2 = 146
<=> 2y^2 - 32y + 110 = 0
<=> y^2 - 16y + 55 = 0
<=> (y-11)(y-5) = 0
<=> y=11 or y= 5
For y = 11, we have x= 16-11=5
For y=5, we have x= 16-5=11
Answer: (x,y) = (11,5) ; (5,11)
Answer:
Step-by-step explanation:
1965 year silver got taken out of quarters
2x + y ≤ 8
2x - 5y < 20
Answer: answer below.
Step-by-step explanation:
To solve this system of linear inequalities, we can use a graphical method or algebraic method.
Let's start with the algebraic method.
First, let's rearrange the inequalities to solve for one variable in terms of the other.
From the first inequality, we have:
x ≤ 10 - 2y
From the second inequality, we have:
y ≤ 8 - 2x
From the third inequality, we have:
x ≤ (20 + 5y)/2
Now, let's plot the graphs of these inequalities on a coordinate plane.
Graphing the first inequality, x ≤ 10 - 2y, we start by drawing the line x = 10 - 2y. Since it is a "less than or equal to" inequality, we will draw a solid line.
Graphing the second inequality, y ≤ 8 - 2x, we start by drawing the line y = 8 - 2x. Again, since it is a "less than or equal to" inequality, we will draw a solid line.
Graphing the third inequality, x ≤ (20 + 5y)/2, we start by drawing the line x = (20 + 5y)/2. This time, since it is a "less than" inequality, we will draw a dashed line.
Now, we shade the region that satisfies all three inequalities. This region is the intersection of the shaded regions of the individual inequalities.
Finally, we can determine the solution by looking at the shaded region on the graph. The solution is the set of all points that lie within or on the boundary of the shaded region.
Alternatively, we can also solve the system of inequalities algebraically by finding the points where the lines intersect. We can then check if these points satisfy all three inequalities.
a. True
b. False