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3 years ago

How to solve using substitution y=3/4X+5 4x-3y=-1

1 answer:

5 0

Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as y=(3/4)X+5

Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:

4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)

Let 's now solve the system

4y=3x + 20
4x-3y=-1

We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:

12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)

Then we have

12y = 9x + 60
-12y =-16x - 4

If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8

We were given equations
y=3/4X+5
4x-3y=-1

We can subst. x=8 into either of these eqn's to find y. Let's try the first one:

y = (3/4)(8)+5 = 6+5=11

Then x=8 and y=11.

You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?

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2 years ago

-1(x+5)=3[x+(2x-1)] solved and justified each step

madreJ [45]

Answer:

Step-by-step explanation:

$-1(x+5)=3[x+2x-1)]\\\\\text{for}\ -1(x+5):\ \text{distribtutive property}\\\text{for}\ [x+(2x-1)]:\ \text{associative property}\\\\(-1)(x)+(-1)(5)=3[(x+2x)-1]\\-x-5=3(3x-1)\\\\\text{for}\ 3(3x-1):\ \text{distributive property}\\\\-x-5=(3)(3x)+(3)(-1)\\-x-5=9x-3\\\\\text{for the equation}:\ \text{addition property of equality}\\\\-x-5=9x-3\qquad\text{add 5 to both sides}\\-x-5+5=9x-3+5\\-x=9x+2\\\\\text{for the equation:}\ \text{subtraction property of equality}\\\\-x=9x+2\qquad\text{subtract}\ 9x\ \text{from both sides}$

$-x-9x=9x-9x+2\\-10x=2\\\\\text{for the equation:}\ \text{division property of equality}\\\\-10x=2\qquad\text{divide both sides by -10}\\\dfrac{-10x}{-10}=\dfrac{2}{-10}\\x=-0.2$

$\text{Distributive property:}\ a(b+c)=ab+ac\\\text{Associative property:}\ (a+b)+c=a+(b+c)$

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2 years ago