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3 years ago

Solve the system using substitution: y + 8x = 10 and 2y - 4x = 40

Mathematics

1 answer:

exis [7]3 years ago

4 0

Answer:

Hence x = -1 and y = 18

Step-by-step explanation:

Given the system of equations

y + 8x = 10... 1

2y - 4x = 40 ... 2

From 1; y = 10-8x

Substitute into 2;

2(10-8x) - 4x = 40

20 - 16x - 4x = 40

20 - 20x = 40

-20x = 40 - 20

-20x = 20

x = -1

Recall that y = 10-8x

y = 10 - 8(-1)

y = 10 + 8

y = 18

Hence x = -1 and y = 18

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evablogger [386]

Answer:

5 1/4

Step-by-step explanation:

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8 0

3 years ago

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an airplaine descends 1.5 miles to an elevation of 5.25 miles. find the elevation of the plane before its descent

Maksim231197 [3]

We can then write an equation representing this problem as:

e−1.5mi=5.25mi

Now, add 1.5mi to each side of the equation to solve for e while keeping the equation balanced:

e−1.5mi+1.5mi=5.25mi+1.5mi

e−0=6.75mi

e=6.75mi

The plane's starting elevation was 6.75 miles

Hope this helps!

8 0

3 years ago

The base of a triangle is seven less than twice its height. If the area of the triangle is 100cm^2 find the length of the base

Alexeev081 [22]

2x-7 * x
----------- =. 100^2= 10000
2

2x^2-7/2= x^2-7/2

x^2-7/2=10000

x^2 = 20007/7

7 0

4 years ago

Read 2 more answers

Pls help me fast and give me the steps pls 1+7p-40=80

valentinak56 [21]

Answer:

Step-by-step explanation:

1+7p-40=80

-39+7p=80 Move constant to the right

7p=80+39 add the numbers

7p=119 divide both sides by 7

4 0

3 years ago

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Х- а x-b If f(x) = b.x-a÷b-a + a.x-b÷a - b Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

5 0

3 years ago