1. since the equations equal each other
2x+19=x+7
-x -x
x+19=7
-19 -19
x= -12
sub into y= x+7
y= -12+7
y= -5
2. x+2y=-7
<span>x+y+23=0
i don't know what the second equation is equal to, i'll assume 0
2y=x-7
y=1/2x-7/2
sub this into x+y+23=0
x+1/2x-7/2+23=0
3/2x+39/2=0
-39/2 -39/2
3/2x= -39/2
x=-39/2 / 3/2
x= -13
sub into y=1/2x-7/2
y=1/2(-13)-7/2
y= -10
3. </span>x+5y=2
<span>x=-4y+5
</span>sub x=-4y+5 into x+5y=2
-4y+5+5y=2
y+5=2
-5 -5
y= -3
sub y =-3 into x =-4y+5
x=-4(-3)+5
x= 17
4. 3x+y=9
<span>y=-5x+9
</span>sub -5x+9 into 3x+y=9
3x+(-5x+9)=9
3x-5x+9=9
-2x+9=9
-9 -9
-2x=0
x=0
sub into y=-5x+9
y=-5(0)+9
y=9
The graph y=|x|-4 is obtained from the graph y=|x| dy <span>moving down 4 units the graph y=|x| along the y-axis (see, if x=0, then for y=|x|, y=0 and for y=|x|-4, y=-4).
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These two graphs have the same form.
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Answer: Any real number x as long as
and 
In other words, anything but 0 or -2/3 is valid.
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Explanation:
Set the denominator equal to zero and solve for x
2(3x^2 + 2x) = 0
3x^2 + 2x = 0
x(3x + 2) = 0
x = 0 or 3x+2 = 0 .... zero product property
x = 0 or 3x = -2
x = 0 or x = -2/3
If either x = 0 or x = -2/3, then the denominator 2(3x^2 + 2x) will be zero. But recall that we cannot have zero in the denominator. Dividing by zero is not allowed. The expression is undefined when we divide by zero.
Therefore, we must exclude x = 0 and x = -2/3 from the domain. Any other real number is valid as an x input.
Assuming the interest is simple interest, as opposed to compound interest,
Yes because 15+8=23 divide that into a number that could each child have and just make it into equal amounts