Question

Check the solution of the following equation [i] 6x+3=2x+12, at x=2 [ii] 9x-2[x+5]=5[x+6], at x=20 Please tell the answer

Answer 1

Answer:

$\huge\underline\mathfrak\red{✨ Answer✨}$

4x2 + 5x2 - 2x + 8x. = 9x2 + 6x. = 3x(3x + 2). Answer: 3x(3x + 2). 2. Simplify: (-7x2 - 3x) - (4x2 - x) Solution: (-7x2 ... -6x + 3 + x2 - 5x = -6x - 5x + x2 + 3 = x2 – 11x + 3. Answer: x2 – 11x + 3 6.

Please let me mark at brainliest.....

one answer least

Answer 2

Answer:

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x Equation 1y, equals, 2, x, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24 Equation 2x, plus, y, equals, 24, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray

The tricky thing is that there are two variables, xxx and yyy. If only we could get rid of one of the variables...

Here's an idea! Equation 111 tells us that \goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 and \goldD yystart color #e07d10, y, end color #e07d10 are equal. So let's plug in \goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 for \goldD yystart color #e07d10, y, end color #e07d10 in Equation 222 to get rid of the yyy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}

x+y

x+2x

=24

=24

Equation 2

Substitute 2x for y

Brilliant! Now we have an equation with just the xxx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3