All categories

2 years ago

-5x - 2y = 13 -4x + y = 0 Solve by substation

Mathematics

1 answer:

kotegsom [21]2 years ago

5 0

Answer:

(x,y)=(-1,-4)

Step-by-step explanation:

-4x + y = 0

+4x +4x

y=4x

-5x - 2y = 13

+5x +5x

-2y = 5x+13

-2(4x) = 5x+13

-8x=5x+13

-5x -5x

-13x/-13=13/-13

x=-1

Thus:

y=4x

y=4(-1)

y=-4

You might be interested in

Exactly 9 years ago, Jeri opened a bank account that earned 2% interest each year. If the account now has $215.12, how much did

MAVERICK [17]

B is the corect aswer

4 0

2 years ago

A pair of parallel lines is cut by a transversal as shown in the figure. Match the angles with their corresponding angles.

garri49 [273]

D and H are corresponding
G and C are corresponding
E and A are corresponding
B and F are corresponding

4 0

2 years ago

Read 2 more answers

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.