All categories

4 years ago

How can you use substitution method to solve a system of equations that does not have a variable with a coefficient of 1 or –1?

1 answer:

6 0

You can multiply or divide one equation so that it has a variable with a coefficient opposite of the other equation.
Example:
A) 2x + 3y = 6
B) 4x +2y = 20
Notice that neither equation has a "1" or "-1"
Let's multiply Equation A) by -2
A) -4x -6y = -12 Then add it to B)
B) 4x +2y = 20
-4y = 8
y= -2

A) 2x + 3*-2 = 6
2x = 12
x = 6

You might be interested in

If J/24 = K, then J/6 =

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

$\frac{J}{24}$ = K ( multiply both sides by 24 to clear the fraction )

J = 24K ( divide both sides by 6 )

$\frac{J}{6}$ = 4K

8 0

3 years ago

What is 0.7 in inches

Ainat [17]

.7 inches? You need to put what to convert it from

8 0

3 years ago

Read 2 more answers

Show all work for full credit. Given: f (x) = -2x + 1, find f (1) x Clear h Undo

xenn [34]

Answer:

I hope this helps

Step-by-step explanation:

five percent

4 0

3 years ago

Multiply 15/8 by 4/9

ddd [48]

The answer is 5/6:) :)

8 0

4 years ago

Read 2 more answers

Please help !!! I can’t solve this The teacher doesn’t teach well

Scorpion4ik [409]

So here are 4 rules when it comes to exponents:

When you raise a power to a power, you really just multiply the exponents: $(x^m)^n=x^{m*n}$
When you multiply 2 exponents of the same base, you add the exponents together: $x^m*x^n=x^{m+n}$
When you divide 2 exponents of the same base, you subtract the exponents: $\frac{x^m}{x^n}=x^{m-n}$
To convert a negative exponent into a positive one, flip it to its reciprocal: $x^{-m}=\frac{1}{x^m}\\ \frac{1}{x^{-m}}=\frac{x^{m}}{1}=x^m$

Part A:

$\frac{2x^3y^3}{4y^2}=\frac{1}{2}x^{3-0}y^{3-2}=\frac{1}{2}x^3y$

Part B:

$(\frac{x^{-8}}{y^{11}})^{-2}=\frac{x^{-8*-2}}{y^{11*-2}}=\frac{x^{16}}{y^{-22}}=\frac{x^{16}y^{22}}{1}=x^{16}y^{22}$

Part C:

$\frac{(2x^3)(x^4)^2}{8x^{11}}=\frac{(2x^3)(x^{4*2})}{8x^{11}}=\frac{(2x^3)(x^8)}{8x^{11}}=\frac{2x^{3+8}}{8x^{11}}=\frac{2x^{11}}{8x^{11}}=\frac{1}{4}x^{11-11}=\frac{1}{4}x^0=\frac{1}{4}$

5 0

3 years ago