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3 years ago

What is the value of y in this system of equations: 3x + y = 9 y = –4x + 10 ?

1 answer:

6 0

3x + y = 9 (I)
y = –4x + 10 (II)
------------------------
 $\left \{ {{3x + y = 9 } \atop {y = -4x + 10 }} \right.$

Pass the incognito "4x" to the first term, changing the signal when changing sides.
-------------------------
simplify by (-1)
 $\left \{ {{3x + y = 9.(-1)} \atop {4x + y = 10 }} \right.$ 

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 $\left \{ {{- 3x - \diagup\!\!\!\!y =-9} \atop {4x + \diagup\!\!\!\!y = 10}} \right. 
$

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 $\left \{ {{- 3x = - 9} \atop {4x = 10}} \right.$ 

-----------
 $\boxed{x = 1}$ 

Substitute in equation (I) to find the value of "Y".

3x + y = 9 (I)
3*(1) + y = 9
3 + y = 9
y = 9 - 3
$\boxed{y = 6}$

Answer:
$\boxed{\boxed{ \left \ {{x=1} \atop {y=6}} \right. }}$

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To solve x in this equation we must first add five to both sides to get $\sqrt{x}$ on its own.

Original Equation :
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New Equation {Added 5 to Both Sides} :
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Now we must square both sides of the equation.

Old Equation :
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New Equation {Changed by Squaring Both Sides} :
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And now we must solve the new equation.

Step 1 - Switch sides
$x^{2} -12x+36=x$

Step 2 - Subtract x from both sides
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Step 3 - Simplify
$x^{2} -13x+36=0$

Now we need to solve the rest of the equation using the quadratic formula.

$\frac{-(-13)+ \sqrt{(-13) ^{2}-4*1*36 } }{2*1}$

${13+ \sqrt{(-13)^{2}-4*1*36 }=13+ \sqrt{25}$

$\frac{13+ \sqrt{25}}{2}$

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$\frac{-(-13)- \sqrt{(-13) ^{2} -4*1*36} }{2*1}$

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So, this means that in the equation $\sqrt{x} -5=x-11$ , x = 9 and x = 4.

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