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2 years ago

2x+4y=60 and y=x+3 i need a answer quick

Mathematics

1 answer:

Kay [80]2 years ago

7 0

It is asking for the intercept.

y = x + 3 becomes y - x = 3

multiply it by two, so it becomes -2x + 2y = 6. Then add it to the other equation, and solve for y.

2x + 4y = 60

(2x + 4y) + (-2x + 2y) = 60 +6

6y = 66

y = 11

Plug in the known y into the the easiest equation.

11 = x +3

x = 8

Intercept: (8, 11)

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Solve the given system of equations utilizing either the substitution or addition method.

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Answer:

Value of x=4 and y=-13/3

Step-by-step explanation:

We need to solve the systems of equations

$5x+3y=7 \ and \ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}$

Solving using substitution method:

Let:

$5x+3y=7---eq(1) \\ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}---eq(2)$

Find value of x from eq(1) and put in eq(2)

$5x+3y=7\\5x=7-3y\\x=\frac{7-3y}{5}$

Put value of x in eq(2)

$\frac{3}{2}(\frac{7-3y}{5}) -\frac{3}{4}y =\frac{37}{4}\\\frac{21-9y}{10} -\frac{3}{4}y =\frac{37}{4}\\Taking \ LCM\\\frac{21*2-9y*2-3y*5}{20} =\frac{37}{4}\\\frac{42-18y-15y}{20} =\frac{37}{4}\\42-33y=\frac{37}{4}*20\\42-33y=185\\-33y=185-42\\-33y=143\\y=\frac{143}{-33}\\y=-\frac{13}{3}$

Now, finding value of x by putting value of y in eq(1)

$x=\frac{7-3y}{5}\\x= \frac{7-3(-\frac{13}{3}) }{5}\\x=\frac{7+13}{5}\\x=\frac{20}{5}\\x=4$

So, Value of x=4 and y=-13/3

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Fie triunghiul ABC isoscel cu AB=AC=3 cm daca mediatoarea laturi AC intersectat cu latura BC in M si perimetrul thriunghiului AM

NemiM [27]

Answer:

MC = 4.5cm

Step-by-step explanation:

Question:

Let the isosceles triangle ABC with AB = AC = 3 cm. if the mediator of the sides AC intersects with the side BC in M and the perimeter of the triangle AMC = 12 cm. Calculate MC.

Solution:

Find attached the diagram used in solving the question.

Given:

∆ABC is an isosceles triangle (two sides and angles are equal)

AB = BC = 3cm

Perimeter of ∆AMC = 12cm

From the diagram, M cuts AC at the the middle.

AD = CD = AC/2 = 3/2

Perimeter of Right angled ∆AMD = AM + AD + MD

= 3/2 + AM +MD

Perimeter of Right angled ∆CMD =CM + CD + MD

= 3/2 + CM +MD