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3 years ago

How do you solve by substitution in algebra?

1 answer:

7 0

Ok so lets say we have a system of equations:

y = x + 1
x + y = 5

All you would do is plug in what is equal to the variable so:

x + x + 1 = 5

Then you would just simplify and solve

2x + 1 = 5
-1 -1
_________

2x = 4
-- --
2 2

x=2

You might be interested in

Solve for x. 8^(x+3)=32

Damm [24]

Answer:

x = -4/3

Step-by-step explanation:

8^(x+3)=32

2^3 = 8

2^5 = 32

3( x + 3 ) = 5

3x + 9 = 5

3x = -4

x = -4/3

Tell me if I am wrong.

Can I get brainliest

4 0

3 years ago

Read 2 more answers

Please help me with my math homework what is the answer!!!

Dmitry [639]

Answer:

Step-by-step explanation:

The letters are easy enough.

y^3 determines that 3 ys are needed. y^2 is taken in by the cubed.

x^4 determines that 4 xs are needed. x^3 is part of x^4 and you don't need any more xs

12 and 42 are the parts that will cause the problem. Factor them both into prime factors

12: 2 * 2 * 3

42: 2 * 3 * 7

You need two 2s.

You need one 3

You need one 7

LCD = 2 *2*3 * 7 = 84

6 0

2 years ago

Write the product of 6x-4 and the multiplicative inverse of 1 over 5

nlexa [21]

Answer:

30x -20

Step-by-step explanation:

The multiplicative inverse (reciprocal) of 1/5 is 5/1 = 5. Multiplying that by 6x-4 gives ...

5 × (6x -4) = 30x -20

6 0

3 years ago

Solve : a) (x-4)(x-2)=0

Elden [556K]

Answer: X= 3

Step-by-step explanation:

x-4*x-2=0

+4. +4

—————-

x*x-2=4

+2. +2

——————

x*x=6

2x=6

/2. /2

—————-

x=3

8 0

3 years ago

An open box is to be constructed so that the length of the base is 3 times larger than the width of the base. If the cost to con

natta225 [31]

Answer:

Step-by-step explanation:

Let assume that the volume = 88 cubic feet

Then:

$L = 3w --- (1)volume = l \times w \times h \\ \\ 88 = (3w) \times w \times h \\ \\ 3w^2 h= 88 \\ \\ h = \dfrac{88}{3w^2}--- (2) \\ \\$

The construction cost now is:

$C = 4(l \times w) + 2 ( w \times h) + 2(l \times h) \\ \\ C = 4(3w^2) + 2(w \times \dfrac{88}{3w^2}) + 2(3w \times \dfrac{88}{3w^2}) \\\\ C = 12w^2 + \dfrac{176}{3w} + \dfrac{176}{w}$

Now, to determine the minimum cost:

$\dfrac{dC}{dw}= 0 \\ \\ \implies 24 w - \dfrac{176}{3w^2}- \dfrac{176}{w^2}=0 \\ \\ 24 w ^3 = \dfrac{176}{w^2}(\dfrac{1}{3}+1) \\ \\ 24 w ^3 = \dfrac{176(4)}{3} \\ \\ w^3 = \dfrac{88}{3(3)}$

$w = \dfrac{88}{3(3)}^{1/3} \ feet$

Now;

$length = 3 ( \dfrac{88}{3(3)})^{1/3} \ feet$

$height = ( 88})^{1/3} (3)^{1/3} \ feet$

6 0

3 years ago