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Kay [80]
3 years ago
15

Using substitution to solve the system below, what expression should be substituted into the first equation? 2x-4y=-4 and x+2y=8

Mathematics
2 answers:
inna [77]3 years ago
5 0

Answer:

Given the system of equation:

2x-4y=-4                    ......[1]

x+2y=8                      ......[2]

we can rewrite equation [2] as;

x = 8-2y                ......[3]

Substitute equation [3] into [1] to eliminate x, and solve for y;

2(8-2y)-4y = -4

Using distributive property: a\cdot (b+c) = a\cdot b+ a\cdot c

16-4y -4y = -4

Combine like terms;

16 - 8y = -4

Add 4 to both sides we have;

20 - 8y = 0

Add 8y to both sides we have;

20 = 8y

Divide 8 to both sides we have;

y = 2.5

Substitute the y-value in [3] we have;

x = 8-2(2.5)

x = 8 - 5 = 3

Therefore, the  expression should be substituted into the first equation is, x = 8-2y  and also the value of x = 3 and y = 2.5

MA_775_DIABLO [31]3 years ago
5 0

Answer:


Step-by-step explanation:

Choices?

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Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

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Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

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