Solve the system by substitution. 3x-6 = y -6x + y = 9

Determine the input value for which the statement f(x) = g(x) is true.

Naya [18.7K]

Answer:

(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻(╯°□°）╯︵ ┻━┻

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$