y=x+4 (y+3)/(4)=7x^(2)-x+1 If (a,b) is a solution to the system of equations shown and a 0, what is the value of a ?

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y=x+4  (y+3)/(4)=7x^(2)-x+1 If  (a,b) is a solution to the system of equations shown and  a > 0, what is the value of  a ?

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Substitute y in second equation: Substitute the expression for y from the first equation into the second equation. Since y = x + 4, we can replace y in the second equation with x + 4. (y + 3)/4 = 7x^2 - x + 1 becomes ((x + 4) + 3)/4 = 7x^2 - x + 1.Simplify left side: Simplify the left side of the equation. ((x + 4) + 3)/4 = (x + 7)/4. Now we have (x + 7)/4 = 7x^2 - x + 1.Multiply by 4: Multiply both sides of the equation by 4 to eliminate the fraction. 4 * ((x + 7)/4) = 4 * (7x^2 - x + 1). This simplifies to x + 7 = 28x^2 - 4x + 4.Rearrange to set to zero: Rearrange the equation to set it to zero and find the values of x. 28x^2 - 4x + 4 - x - 7 = 0. This simplifies to 28x^2 - 5x - 3 = 0.Solve quadratic equation: Solve the quadratic equation for x. This is a standard quadratic equation in the form ax^2 + bx + c = 0. We can use the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), to find the values of x. Here, a = 28, b = -5, and c = -3.Calculate discriminant: Calculate the discriminant (b^2 - 4ac) to determine the nature of the roots. Discriminant = (-5)^2 - 4 * 28 * (-3). Discriminant = 25 + 336. Discriminant = 361.Calculate roots: Since the discriminant is positive, we have two real roots. Calculate the roots using the quadratic formula. x = (-(-5) ± √361) / (2 * 28). x = (5 ± 19) / 56.Find possible values for x: Find the two possible values for x. x = (5 + 19) / 56 or x = (5 - 19) / 56. x = 24 / 56 or x = -14 / 56. x = 3/7 or x = -1/4.Choose positive value for x: Since we are given that a > 0 and a corresponds to the x-value in the solution (a,b), we choose the positive value for x. Therefore, a = 3/7.

$y=x+4$ $\newline$ $\frac{y+3}{4}=7 x^{2}-x+1$ $\newline$ If $(a, b)$ is a solution to the system of equations shown and a>0 , what is the value of $a$ ?

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y=x+4 y=x+4 y=x+4\newliney+34=7x2−x+1 \frac{y+3}{4}=7 x^{2}-x+1 4y+3​=7x2−x+1\newlineIf (a,b) (a, b) (a,b) is a solution to the system of equations shown and a>0 , what is the value of a a a ?

$y=x+4$ $\newline$ $\frac{y+3}{4}=7 x^{2}-x+1$ $\newline$ If $(a, b)$ is a solution to the system of equations shown and a>0 , what is the value of $a$ ?