X-4=10;4x+1=57i need to change the first equation into to the second equation.

Mathematics

1 answer:

cestrela7 [59]3 years ago

4 0

Is this what you mean this is what i understood.

X-4=10;4x+1=57

First equation solve for x.

X-4=10

Add 4 both sides

X=10+4

X=14

Second equation substitute your x answer into second equation to make both sides equal.

4x+1=57

4(14)+1=57

56+1=57

57=57

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Solve 5x^2 - 7x + 2 = 0 by completing the square.

Reika [66]

Answer: The correct option is (c) $\dfrac{49}{100}.$

Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:

$5x^2-7x+2=0~~~~~~~~~~~~~~~~~~~(i)$

Also, we are to find the constant added on both sides to form the perfect square trinomial.

We have from equation (i) that

$5x^2-7x+2=0\\\\\Rightarrow x^2-\dfrac{7}{5}x+\dfrac{2}{5}=0\\\\\\\Rightarrow x^2-2\times x\times \dfrac{7}{10}+\left(\dfrac{7}{10}\right)^2+\dfrac{2}{5}=\left(\dfrac{7}{10}\right)^2\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49}{100}-\dfrac{2}{5}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49-40}{100}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{9}{100}\\\\\\\Rightarrow x-\dfrac{7}{10}=\pm\dfrac{3}{10}\\\\\\\Rightarrow x=\pm\dfrac{3}{10}+\dfrac{7}{10}.$

So,

$x=\dfrac{3}{10}+\dfrac{7}{10},~~~~~~~~x=-\dfrac{3}{10}+\dfrac{7}{10}\\\\\\\Rightarrow x=\dfrac{10}{10},~~~~~~~~\Rightarrow x=\dfrac{-3+7}{10}\\\\\\\Rightarrow x=1,~-\dfrac{2}{5}.$