5x+3y=-6 ...(1)
2x-3y=27 ...(2)
We can add equations (1) and (2) to eliminate y . So,
7x+0=21
7x=21
Divide each sides by 21 to isolate x.
So, x= 3
Now plug in x=3 in equation (1) to get the value of y.
So, 5(3)+3y=-6
15 +3y= -6
15+3y-15=-6-15
3y= -21
So, y=-7.
Hence, x=3 and y=7.
Let x = number of quarters
since she has twice as many dimes as quarters,
Let 2x = number of dimes
The total value in $ is
0.25x + (0.10)(2x)
And it was said that when the numbers of quarters and dimes are interchanged, she would have $1.20 more than the original. So,
(New Value) = (Original Value) + $1.20
0.25(2x) + 0.10x = 0.25x + (0.10)(2x) + 1.20
Solving for x,
0.5x + 0.1x = 0.25x + 0.2x + 1.20
0.6x = 0.45x + 1.20
0.6x - 0.45x = 1.20
0.15x = 1.20
x = 1.20/0.15
x = 8 quarters
2x = 16 dimes
Hey there! :D
Just plug in the values of x.
y= 1/2(-5)-2
y= -2.5-2
y=-4.5
First point: (-5, -4.5)
y= 1/2(-2)-2
y=-1-2
y=-3
Second point: (-2, -3)
y= 1/2(1)-2
y =.5-2
y=-1.5
Third point: (1,-1.5)
y= 1/2(3)-2
y= 1.5-2
y=-.5
Fourth point: (3, -0.5)
y= 1/2(6)-2
y=3-2
y=1
Fifth point: (6,1)
I hope this helps!
~kaikers