You want to eliminate one of the terms (x or y) in one of the equations so you can solve for the other variable. You have to multiply by the opposite number of the coefficient to be able to eliminate the term in the other equation. If the x coefficient is 2, then you have to multiply the entire other equation by -2. If the y coefficient is -5, then you have to multiply the entire other equation by 5.
10)
-4x + 9y= 9
x - 3y= -6
STEP 1:
multiply the bottom equation by 4
4(x- 3y)= 4(-6)
4x - 12y= -24
STEP 2:
add the top equation and the equation from step 2
-4x + 9y= 9
4x - 12y= -24
the x term cancels out
-3y= 15
divide both sides by -3
y= -5
STEP 2:
substitute the y value in either original equation to solve for x
x - 3y= -6
x - 3(-5)= -6
x + 15= -6
subtract 15 from both sides
x= -21
ANSWER: x= -21; y= -5
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12)
-7x + y= -19
-2x + 3y= -19
STEP 1:
multiply the top equation by -3 to eliminate the y term and to solve for x
-3(-7x + y)= -3(-19)
21x - 3y= 57
STEP 2:
add the bottom equation and the equation from step 2 to solve for x
-2x + 3y= -19
21x - 3y= 57
the y term cancels out
19x= 38
divide both sides by 19
x= 2
STEP 3:
substitute the x value in step 2 to solve for y; you can use either original equation
-7x + y= -19
-7(2) + y= -19
-14 + y= -19
add 14 to both sides
y= -5
ANSWER: x= 2; y=-5
Hope this helps! :)
Answer: The correct answer is (B)
Step-by-step explanation: No. Ari should have written the percent ratio as 5/100
Sub (x+1) for x
f(x+1)=2(x+1)^2+3
f(x+1)=2(x^2+2x+1)+3
f(x+1)=2x^2+4x+2+3
f(x+1)=2x^2+4x+5
D is answer
Y=1.22x
y is the cost and x is kilometers driven. I got 1.22 by dividing 366 by 300 to find how it costs per kilometer.
<h3>Question:</h3>
2x-5y=-28 find the value of y when x equals -19
<h3>Answer:</h3>
Y = - 2
<h3>Step-by-step explanation:</h3>
First you plug in -19 where x is 2(-19)-5y = -28
Remove parentheses -2 · 19-5y= -28
Multiply -2 by 19 -38-5y= -28
Add 38 to both sides -5y = 10
Divide both sides by -5 [tex]\frac{-5y}{-5} =\frac{10}{-5}[/tex
you will get -2 which is the answer.
Hope this helped!
