Solving systems of equations by substitution Y=6x y=5x+7

2 answers:

8 0

Y=6x (1)
y=5x-7 (2)

Substitute y into (2)
(6x)=5x-7 -- subtract 5x from both sides
x=-7

Sub x into 1
y=6(-7)
y=-42

x=-7
y=-42

dusya [7]3 years ago

6 0

First substitute 'y' in 2nd equation by right side of 1st one and you get:
$Y=6x$
$6x=5x+7$
now substract 5x from both sides of 2nd equal and you get:
$6x-5x=5x+7-5x$
as a result of the equal above you get:
$x=7$
now substitute x in very first equation by the result from above (x=7):
$Y=6*7=42$
$x=7$

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_____

There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.