1. What is the solution to the system below? x+y=5 1x- y=3

Mathematics

1 answer:

Alika [10]3 years ago

5 0

Answer:

(4,1)

Step-by-step explanation:

x+y=5

x- y=3

Add the equations together to eliminate y

x+y=5

x- y=3

--------------------

2x = 8

Divide each side by 2

2x/2 = 8/2

x = 4

Now we can find y

x+y = 5

4+y = 5

Subtract 4 from each side

4+y-4 = 5-4

y =1

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The difference between the roots of the equation 3x2+bx+10=0 is equal to 4 1 3 . find<br> b.

inna [77]

Given that the difference between the roots of the equation $3x^2+bx+10=0$ is $4 \frac{1}{3} = \frac{13}{3}$ .

Recall that the sum of roots of a quadratic equation is given by $- \frac{b}{a}$ .

Let the two roots of the equation be $\alpha$ and $\alpha + \frac{13}{3}$ , then

$\alpha + \alpha + \frac{13}{3} =2 \alpha + \frac{13}{3} =- \frac{b}{a} =- \frac{b}{3} \\ \\ i.e.\ \ 2 \alpha + \frac{13}{3}=- \frac{b}{3}$ . . . (1)

Also recall that the product of the two roots of a quadratic equation is given by $\frac{c}{a}$ , thus:

$\alpha \left( \alpha + \frac{13}{3} \right)= \alpha ^2+ \frac{13}{3} \alpha = \frac{c}{a} = \frac{10}{3} \\ \\ i.e.\ \ \alpha ^2+ \frac{13}{3} \alpha=\frac{10}{3}$ . . . (2)

From (1), we have:

$2 \alpha =- \frac{b}{3} - \frac{13}{3} \\ \\ \Rightarrow \alpha =- \frac{b}{6} - \frac{13}{6}$

Substituting for alpha into (2), gives:

$\left(- \frac{b}{6} - \frac{13}{6}\right)^2+ \frac{13}{3} \left(- \frac{b}{6} - \frac{13}{6}\right)= \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} + \frac{13b}{18} + \frac{169}{36} - \frac{13b}{18} - \frac{169}{18} = \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} - \frac{289}{36} =0 \\ \\ \Rightarrow \frac{b^2}{36} = \frac{289}{36} \\ \\ \Rightarrow b^2=289 \\ \\ \Rightarrow b=\pm\sqrt{289}=\pm17$

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3 years ago

Hi I need help with this problem. can anyone you tell me what 5+(-3)

Tomtit [17]

The answer for 5+(-3) is 2

8 0

2 years ago

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