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2 years ago

2x+y=3, 5x-2y=12 how to solve this

2 answers:

7 0

Start by getting one of the equations in terms of one variable, in this case lets put the first equation in terms of y

2x+y=3 ---> y=3-2x

now we plug that into the second equation

5x-2y=12 ---> 5x-2(3-2x)=12

we can then multiply and simplify

5x-2(3-2x)=12 ---> 5x-6+4x=12 ---> 9x-6=12 ---> 9x=18 ---> x=2

we can now plug x=2 into an equation to solve for y

2(2)+y=3 ---> 4+y=3 ---> y=-1

Brums [2.3K]2 years ago

7 0

2x+y=3 =4x+2y=6
5x-2y=12. =5x-2y=12
--------------------------------
9x=18: x=2
4+y=3:y=-1

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What equation represents the line that passes through points (1.-5) and (3 -17)?

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ALL LINES CAN BE REPRESENTED IN THE FORM $f(x) = ax+b$ .

We know that:

$f(1) = -5 \implies a\cdot(1)+b = -5 \implies a+b = -5$

and

$f(3) = -17 \implies a\cdot(3) +b = -17 \implies 3a+b = 17$

Observe that:

$3a+b = 2a+a+b = 2a+(a+b)$

So:

$2a+(a+b) =-17$

But we know that $a+b = -5$ , so:

$2a+(a+b) =-17$

$2a+(-5) = -17$

$2a = -12$

$a = -6$

Since:

$a+b = -5$

$(-6)+b = -5$

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3 years ago

Express in Simper Fom : y-x

AveGali [126]

Simplify

$E=\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}$

Conditions for the existance of $E:$

$x\neq 0~~\text{ and}~~y\neq 0~~\text{ and }~~x\neq -y.$

(otherwise, the one of the denominators would be zero.)
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Simplifying:

$E=\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}$

Multiply both the numerator and the denominator by $xy:$

$E=\dfrac{xy\cdot \left(\frac{x}{y}-\frac{y}{x} \right )}{xy\cdot \left(\frac{1}{y}+\frac{1}{x} \right )}\\\\\\ E=\dfrac{\frac{xy\cdot x}{y}-\frac{xy\cdot y}{x}}{\frac{xy}{y}+\frac{xy}{x}}\\\\\\ E=\dfrac{x^{2}-y^{2}}{x+y}\\\\\\ E=\dfrac{(x+y)(x-y)}{x+y}$

Simplify common factors:

$E=x-y\\\\\\ \therefore~~\boxed{\begin{array}{c}\dfrac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{y}+\frac{1}{x}}=x-y \end{array}}$

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