4y^(2)+9=6x+3 4y=2x+1 If (x,y) is the solution to the system of equations shown, what is the value of y ? How do I enter a student-produced response on the SAT? [Show me!]S

Solve for y: First, let's solve the second equation for y. 4y = 2x + 1 To find y, we divide both sides of the equation by 4. y = (2x + 1) / 4Substitute into first equation: Now, let's substitute the expression for y into the first equation. The first equation is 4y^(2) + 9 = 6x + 3. Substituting y with (2x + 1) / 4, we get: 4[(2x + 1) / 4]^(2) + 9 = 6x + 3Simplify equation: Next, we simplify the equation by squaring the term (2x + 1) / 4 and multiplying by 4. [(2x + 1) / 4]^(2) = (2x + 1)^(2) / 16 4 * (2x + 1)^(2) / 16 + 9 = 6x + 3 (2x + 1)^(2) / 4 + 9 = 6x + 3Correct previous mistake: Now, we simplify the left side of the equation by multiplying out the square and dividing by 4. (2x + 1)^(2) = 4x^(2) + 4x + 1 (4x^(2) + 4x + 1) / 4 + 9 = 6x + 3 x^(2) + x + 1/4 + 9 = 6x + 3We made a mistake in the previous step by incorrectly simplifying the square of (2x + 1). Let's correct this error. (2x + 1)^(2) = 4x^(2) + 4x + 1 (4x^(2) + 4x + 1) / 4 + 9 = 6x + 3 x^(2) + x + 1/4 + 36/4 = 6x + 3 x^(2) + x + 37/4 = 6x + 3

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4y^(2)+9=6x+3

4y=2x+1
If
(x,y) is the solution to the system of equations shown, what is the value of
y ?
How do I enter a student-produced response on the SAT? [Show me!]S

$4 y^{2}+9=6 x+3$ $\newline$ $4 y=2 x+1$ $\newline$ If $(x, y)$ is the solution to the system of equations shown, what is the value of $y$ ? $\newline$ How do I enter a student-produced response on the SAT? [show me!] S

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Algebra 1

Find the number of solutions to a system of equations

Full solution

4 y^{2}+9=6 x+3

\newline

4 y=2 x+1

\newline

(x, y)

is the solution to the system of equations shown, what is the value of

y

\newline

How do I enter a student-produced response on the SAT? [show me!] S

Solve for y: First, let's solve the second equation for y. $\newline$ $4y = 2x + 1$ $\newline$ To find y, we divide both sides of the equation by $4$ . $\newline$ $y = \frac{2x + 1}{4}$
Substitute into first equation: Now, let's substitute the expression for $y$ into the first equation. $\newline$ The first equation is $4y^{2} + 9 = 6x + 3$ . $\newline$ Substituting $y$ with $(2x + 1) / 4$ , we get: $\newline$ $4[(2x + 1) / 4]^{2} + 9 = 6x + 3$
Simplify equation: Next, we simplify the equation by squaring the term $(2x + 1) / 4$ and multiplying by $4$ . $\left(\frac{2x + 1}{4}\right)^{2} = \frac{(2x + 1)^{2}}{16}$ $4 \cdot \frac{(2x + 1)^{2}}{16} + 9 = 6x + 3$ $\frac{(2x + 1)^{2}}{4} + 9 = 6x + 3$
Correct previous mistake: Now, we simplify the left side of the equation by multiplying out the square and dividing by $4$ . $\newline$ $(2x + 1)^{2} = 4x^{2} + 4x + 1$ $\newline$ $(4x^{2} + 4x + 1) / 4 + 9 = 6x + 3$ $\newline$ $x^{2} + x + 1/4 + 9 = 6x + 3$
Correct previous mistake: Now, we simplify the left side of the equation by multiplying out the square and dividing by $4$ . $\newline$ $(2x + 1)^{2} = 4x^{2} + 4x + 1$ $\newline$ $(4x^{2} + 4x + 1) / 4 + 9 = 6x + 3$ $\newline$ $x^{2} + x + 1/4 + 9 = 6x + 3$ We made a mistake in the previous step by incorrectly simplifying the square of $(2x + 1)$ . Let's correct this error. $\newline$ $(2x + 1)^{2} = 4x^{2} + 4x + 1$ $\newline$ $(4x^{2} + 4x + 1) / 4 + 9 = 6x + 3$ $\newline$ $x^{2} + x + 1/4 + 36/4 = 6x + 3$ $\newline$ $x^{2} + x + 37/4 = 6x + 3$